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][1] 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.


  [1]: http://www.ems-ph.org/journals/show_abstract.php?issn=0013-6018&vol=64&iss=4&rank=4