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It's possible to completely characterize when a six-tuple $(a, b, c, d, e, f)$ forms the lengths of edges of a (non-degenerate) tetrahedron. Namely, by a quick google search I found for example a paper Edge lengths determining tetrahedrons by Wirth and Dreiding, Elem. Math. 64 (2009) 160 – 170.

By their Lemma 2.1, such a 6-tuple (taken as determining not only the lengths of the edges, but also the positions of the edges in the tetrahedron) actually comes from a tetrahedron if and only if each face triangle satisfies triangle inequality and there is a vertex such that the 3 angles on faces around it are all acute and satisify the triangle inequality.

Now we can try to check that if $(a, b, c, d, e, f)$ come from a tetrahedron, then so does $(a+1, b+1, c+1, d+1, e+1, f+1)$: for the faces, just note that adding 1 to the edge lengths makes the triangle more equilateral, and so the triangle inequality will again hold (if $a < b+c$, then $(a+1) < (b+1)+(c+1)$).

EDIT: Ok, for the angles it's more tricky, as Igor Rivin says in his comment.

First, seeing that acute angles remain acute is easy, just from the fact that the angle opposite side $x$ in a triangle with sides $x, y, z$ is acute iff $x^2 < y^2+z^2$ (by the cosine rule).

For the rest we'll need some lower bound on the lengths of edges, the statement in fact does not hold with short edges.

Here is a counterexample: Label the edge lengths so that the faces are $abc$, $dec$, $dbf$ and $aef$. We'll be looking at the vertex $V$ where $a, b, f$ meet. Choose some small $\varepsilon$ and set $a=b=\varepsilon$, $c=\varepsilon^2$, and choose $d, e, f$ very large ($>\varepsilon^{-1}$) so that the angles at $V$ in triangles $dbf$ and $aef$ are small (and the edges come from a tetrahedron). However, after I increase all edge lenths by 1, the triangles $(d+1),(b+1),(f+1)$ and $(a+1),(e+1),(f+1)$ have almost the same shape as $dbf$ and $aef$, hence their angles at $V$ have almost not changed. But the triangle $(a+1),(b+1),(c+1)$ is now almost equilateral, and so its angle at $V$ is almost $\pi/3$. Thus the triangle inequality between angles at $V$ has no chance of holding.

It seems intuitively clear that if we assume that all the sides are $\geq C$ for a constant $C$, than the bigger $C$ is, the less will the angles change when we enlarge the sides by 1. But to make this into a proof of the statement, we'd need some quantitative estimate on how much the angles can change depending on the size of the angle (and $C$). It just seems to me like very technical and not very enjoyable calculations, so I'll be happy to leave them to someone else :) At least, then the solution of the integral problem would follow just by checking the finitely many remaining cases.

It's possible to completely characterize when a six-tuple $(a, b, c, d, e, f)$ forms the lengths of edges of a (non-degenerate) tetrahedron. Namely, by a quick google search I found for example a paper Edge lengths determining tetrahedrons by Wirth and Dreiding, Elem. Math. 64 (2009) 160 – 170.

By their Lemma 2.1, such a 6-tuple (taken as determining not only the lengths of the edges, but also the positions of the edges in the tetrahedron) actually comes from a tetrahedron if and only if each face triangle satisfies triangle inequality and there is a vertex such that the 3 angles on faces around it are all acute and satisify the triangle inequality.

Now we can try to check that if $(a, b, c, d, e, f)$ come from a tetrahedron, then so does $(a+1, b+1, c+1, d+1, e+1, f+1)$: for the faces, just note that adding 1 to the edge lengths makes the triangle more equilateral, and so the triangle inequality will again hold (if $a < b+c$, then $(a+1) < (b+1)+(c+1)$).

EDIT: Ok, for the angles it's more tricky, as Igor Rivin says in his comment.

First, seeing that acute angles remain acute is easy, just from the fact that the angle opposite side $x$ in a triangle with sides $x, y, z$ is acute iff $x^2 < y^2+z^2$ (by the cosine rule).

As Anton Petrunin pointed out, the counterexample was nonsense. Sorry. For the rest we'll need some lower bound on the lengths of edges, the statement in fact does not hold with short edges. 

Here is a counterexample: Label the edge lengths so that the faces are $abc$, $dec$, $dbf$ and $aef$. We'll be looking at the vertex $V$ where $a, b, f$ meet. Choose some small $\varepsilon$ and set $a=b=\varepsilon$, $c=\varepsilon^2$, and choose $d, e, f$ very large ($> \varepsilon^{-1}$) so that the angles at $V$ in triangles $dbf$ and $aef$ are small (and the edges come from a tetrahedron). However, after I increase all edge lenths by 1, the triangles $(d+1),(b+1),(f+1)$ and $(a+1),(e+1),(f+1)$ have almost the same shape as $dbf$ and $aef$, hence their angles at $V$ have almost not changed. But the triangle $(a+1),(b+1),(c+1)$ is now almost equilateral, and so its angle at $V$ is almost $\pi/3$. Thus the triangle inequality between angles at $V$ has no chance of holding. 

It seems intuitively clear that if we assume that all the sides are $\geq C$ for a constant $C$, than the bigger $C$ is, the less will the angles change when we enlarge the sides by 1. But to make this into a proof of the statement, we'd need some quantitative estimate on how much the angles can change depending on the size of the angle (and $C$). It just seems to me like very technical and not very enjoyable calculations, so I'll be happy to leave them to someone else :) At least, then the solution of the integral problem would follow just by checking the finitely many remaining cases.

It's possible to completely characterize when a six-tuple $(a, b, c, d, e, f)$ forms the lengths of edges of a (non-degenerate) tetrahedron. Namely, by a quick google search I found for example a paper Edge lengths determining tetrahedrons by Wirth and Dreiding, Elem. Math. 64 (2009) 160 – 170.

By their Lemma 2.1, such a 6-tuple (taken as determining not only the lengths of the edges, but also the positions of the edges in the tetrahedron) actually comes from a tetrahedron if and only if each face triangle satisfies triangle inequality and there is a vertex such that the 3 angles on faces around it are all acute and satisify the triangle inequality.

Now we can try to check that if $(a, b, c, d, e, f)$ come from a tetrahedron, then so does $(a+1, b+1, c+1, d+1, e+1, f+1)$: for the faces, just note that adding 1 to the edge lengths makes the triangle more equilateral, and so the triangle inequality will again hold (if $a < b+c$, then $(a+1) < (b+1)+(c+1)$). Similarly for the angles (well, this would probably require some proof, probably just checking by a calculation after you express the angles from the edge lengths - which would be messy, but doable... although it may be easier just to compute the determinant from their Theorem 3.1). So if I'm not missing anything, the integer length condition shouldn't play any role.

Edit: Ok, for the angles it's more tricky, as Igor Rivin says in his comment.

First, seeing that acute angles remain acute is easy, just from the fact that the angle opposite side $x$ in a triangle with sides $x, y, z$ is acute iff $x^2 < y^2+z^2$ (by the cosine rule).

For the rest we'll need some lower bound on the lengths of edges, the statement in fact does not hold with short edges.

Here is a counterexample: Label the edge lengths so that the faces are $abc$, $dec$, $dbf$ and $aef$. We'll be looking at the vertex $V$ where $a, b, f$ meet. Choose some small $\varepsilon$ and set $a=b=\varepsilon$, $c=\varepsilon^2$, and choose $d, e, f$ very large ($>\varepsilon^{-1}$) so that the angles at $V$ in triangles $dbf$ and $aef$ are small (and the edges come from a tetrahedron). However, after I increase all edge lenths by 1, the triangles $(d+1),(b+1),(f+1)$ and $(a+1),(e+1),(f+1)$ have almost the same shape as $dbf$ and $aef$, hence their angles at $V$ have almost not changed. But the triangle $(a+1),(b+1),(c+1)$ is now almost equilateral, and so its angle at $V$ is almost $\pi/3$. Thus the triangle inequality between angles at $V$ has no chance of holding.

It seems intuitively clear that if we assume that all the sides are $\geq C$ for a constant $C$, than the bigger $C$ is, the less will the angles change when we enlarge the sides by 1. But to make this into a proof of the statement, we'd need some quantitative estimate on how much the angles can change depending on the size of the angle (and $C$). It just seems to me like very technical and not very enjoyable calculations, so I'll be happy to leave them to someone else :) At least, then the solution of the integral problem would follow just by checking the finitely many remaining cases.

It's possible to completely characterize when a six-tuple $(a, b, c, d, e, f)$ forms the lengths of edges of a (non-degenerate) tetrahedron. Namely, by a quick google search I found for example a paper Edge lengths determining tetrahedrons by Wirth and Dreiding, Elem. Math. 64 (2009) 160 – 170.

By their Lemma 2.1, such a 6-tuple (taken as determining not only the lengths of the edges, but also the positions of the edges in the tetrahedron) actually comes from a tetrahedron if and only if each face triangle satisfies triangle inequality and there is a vertex such that the 3 angles on faces around it are all acute and satisify the triangle inequality.

Now we can try to check that if $(a, b, c, d, e, f)$ come from a tetrahedron, then so does $(a+1, b+1, c+1, d+1, e+1, f+1)$: for the faces, just note that adding 1 to the edge lengths makes the triangle more equilateral, and so the triangle inequality will again hold (if $a < b+c$, then $(a+1) < (b+1)+(c+1)$).

EDIT: Ok, for the angles it's more tricky, as Igor Rivin says in his comment.

First, seeing that acute angles remain acute is easy, just from the fact that the angle opposite side $x$ in a triangle with sides $x, y, z$ is acute iff $x^2 < y^2+z^2$ (by the cosine rule).

For the rest we'll need some lower bound on the lengths of edges, the statement in fact does not hold with short edges.

Here is a counterexample: Label the edge lengths so that the faces are $abc$, $dec$, $dbf$ and $aef$. We'll be looking at the vertex $V$ where $a, b, f$ meet. Choose some small $\varepsilon$ and set $a=b=\varepsilon$, $c=\varepsilon^2$, and choose $d, e, f$ very large ($>\varepsilon^{-1}$) so that the angles at $V$ in triangles $dbf$ and $aef$ are small (and the edges come from a tetrahedron). However, after I increase all edge lenths by 1, the triangles $(d+1),(b+1),(f+1)$ and $(a+1),(e+1),(f+1)$ have almost the same shape as $dbf$ and $aef$, hence their angles at $V$ have almost not changed. But the triangle $(a+1),(b+1),(c+1)$ is now almost equilateral, and so its angle at $V$ is almost $\pi/3$. Thus the triangle inequality between angles at $V$ has no chance of holding.

It seems intuitively clear that if we assume that all the sides are $\geq C$ for a constant $C$, than the bigger $C$ is, the less will the angles change when we enlarge the sides by 1. But to make this into a proof of the statement, we'd need some quantitative estimate on how much the angles can change depending on the size of the angle (and $C$). It just seems to me like very technical and not very enjoyable calculations, so I'll be happy to leave them to someone else :) At least, then the solution of the integral problem would follow just by checking the finitely many remaining cases.

It's possible to completely characterize when a six-tuple $(a, b, c, d, e, f)$ forms the lengths of edges of a (non-degenerate) tetrahedron. Namely, by a quick google search I found for example a paper Edge lengths determining tetrahedrons by Wirth and Dreiding, Elem. Math. 64 (2009) 160 – 170.

By their Lemma 2.1, such a 6-tuple (taken as determining not only the lengths of the edges, but also the positions of the edges in the tetrahedron) actually comes from a tetrahedron if and only if each face triangle satisfies triangle inequality and there is a vertex such that the 3 angles on faces around it are all acute and satisify the triangle inequality.

Now it's easy to check that if $(a, b, c, d, e, f)$ come from a tetrahedron, then so does $(a+1, b+1, c+1, d+1, e+1, f+1)$: for the faces, just note that adding 1 to the edge lengths makes the triangle more equilateral, and so the triangle inequality will again hold. Similarly for the angles (well, this would probably require some proof, probably just checking by a calculation after you express the angles from the edge lengths - which would be messy, but doable... although it may be easier just to compute the determinant from their Theorem 3.1). So if I'm not missing anything, the integer length condition shouldn't play any role.

Edit: Ok, for the angles it's more tricky, as Igor Rivin says in his comment.

First, seeing that acute angles remain acute is easy, just from the fact that the angle opposite side $x$ in a triangle with sides $x, y, z$ is acute iff $x^2 < y^2+z^2$ (by the cosine rule).

For the rest we'll need some lower bound on the lengths of edges, the statement in fact does not hold with short edges.

Here is a counterexample: Label the edge lengths so that the faces are $abc$, $dec$, $dbf$ and $aef$. We'll be looking at the vertex $V$ where $a, b, f$ meet. Choose some small $\varepsilon$ and set $a=b=\varepsilon$, $c=\varepsilon^2$, and choose $d, e, f$ very large ($>\varepsilon^{-1}$) so that the angles at $V$ in triangles $dbf$ and $aef$ are small and the edges form a tetrahedron. However, after I increase all edge lenths by 1, the triangles $(d+1),(b+1),(f+1)$ and $(a+1),(e+1),(f+1)$ have almost the same shape as $dbf$ and $aef$, hence their angles at $V$ have almost not changed. But the triangle $(a+1),(b+1),(c+1)$ is now almost equilateral, and so its angle at $V$ is almost $\pi/3$. Thus the triangle inequality between angles at $V$ has no chance of holding.

It seems intuitively clear that if we assume that all the sides are $\geq C$ for a constant $C$, than the bigger $C$ is, the less will the angles change when we enlarge the sides by 1. But to make this into a proof of the statement, we'd need some quantitative estimate on how much the angles can change depending on the size of the angle (and $C$). It just seems to me like very technical and not very enjoyable calculations, so I'll be happy to leave them to someone else :) At least, then the solution of the integral problem would follow just by checking the finitely many remaining cases.

It's possible to completely characterize when a six-tuple $(a, b, c, d, e, f)$ forms the lengths of edges of a (non-degenerate) tetrahedron. Namely, by a quick google search I found for example a paper Edge lengths determining tetrahedrons by Wirth and Dreiding, Elem. Math. 64 (2009) 160 – 170.

By their Lemma 2.1, such a 6-tuple (taken as determining not only the lengths of the edges, but also the positions of the edges in the tetrahedron) actually comes from a tetrahedron if and only if each face triangle satisfies triangle inequality and there is a vertex such that the 3 angles on faces around it are all acute and satisify the triangle inequality.

Now we can try to check that if $(a, b, c, d, e, f)$ come from a tetrahedron, then so does $(a+1, b+1, c+1, d+1, e+1, f+1)$: for the faces, just note that adding 1 to the edge lengths makes the triangle more equilateral, and so the triangle inequality will again hold (if $a < b+c$, then $(a+1) < (b+1)+(c+1)$). Similarly for the angles (well, this would probably require some proof, probably just checking by a calculation after you express the angles from the edge lengths - which would be messy, but doable... although it may be easier just to compute the determinant from their Theorem 3.1). So if I'm not missing anything, the integer length condition shouldn't play any role.

Edit: Ok, for the angles it's more tricky, as Igor Rivin says in his comment.

First, seeing that acute angles remain acute is easy, just from the fact that the angle opposite side $x$ in a triangle with sides $x, y, z$ is acute iff $x^2 < y^2+z^2$ (by the cosine rule).

For the rest we'll need some lower bound on the lengths of edges, the statement in fact does not hold with short edges.

Here is a counterexample: Label the edge lengths so that the faces are $abc$, $dec$, $dbf$ and $aef$. We'll be looking at the vertex $V$ where $a, b, f$ meet. Choose some small $\varepsilon$ and set $a=b=\varepsilon$, $c=\varepsilon^2$, and choose $d, e, f$ very large ($>\varepsilon^{-1}$) so that the angles at $V$ in triangles $dbf$ and $aef$ are small (and the edges come from a tetrahedron). However, after I increase all edge lenths by 1, the triangles $(d+1),(b+1),(f+1)$ and $(a+1),(e+1),(f+1)$ have almost the same shape as $dbf$ and $aef$, hence their angles at $V$ have almost not changed. But the triangle $(a+1),(b+1),(c+1)$ is now almost equilateral, and so its angle at $V$ is almost $\pi/3$. Thus the triangle inequality between angles at $V$ has no chance of holding.

It seems intuitively clear that if we assume that all the sides are $\geq C$ for a constant $C$, than the bigger $C$ is, the less will the angles change when we enlarge the sides by 1. But to make this into a proof of the statement, we'd need some quantitative estimate on how much the angles can change depending on the size of the angle (and $C$). It just seems to me like very technical and not very enjoyable calculations, so I'll be happy to leave them to someone else :) At least, then the solution of the integral problem would follow just by checking the finitely many remaining cases.