This problem turned out to be much more interesting than I originally
thought. Let me give my solution, which seems to be slightly different from
(but essentially the same as) the solution in the paper by Bremner and
MacLeod (see Allan MacLeod's answer).
Theorem. Let $a,b,c$ be positive integers. Then
$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$ can never be an odd integer.
Let $n$ be a positive odd integer. The equation
$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = n$ implies
$$a^3 + b^3 + c^3 + abc - (n-1)(a+b)(b+c)(c+a) = 0.$$
This describes a smooth cubic curve $E_n$ in the projective plane
that has at least six rational points (of the form $(1:-1:0)$
and $(1:-1:1)$ and their cyclic permutations). Declaring one of
these to be the origin, $E_n$ is an elliptic curve over $\mathbb Q$.
Bringing $E_n$ in Weierstrass form, we obtain the isomorphic curve
$$E'_n \colon y^2 = x \bigl(x^2 + (4n(n+3)-3)x + 32(n+3)\bigr)
=: x(x^2 + Ax + B).$$
If $n = 1$, then there are obviously no positive solutions, so we
assume $n \ge 3$. Then $E_n(\mathbb R)$ has two connected components,
one of which contains the six `trivial' points but no points with
positive coordinates, whereas the other component does contain positive
points. In the model $E'_n$, this component consists of points with
negative $x$-coordinate.
Claim. If $(\xi,\eta) \in E'_n(\mathbb Q)$, then $\xi \ge 0$.
This clearly implies the statement of the theorem.
To show the claim, let $D = 2n + 5$. Then $D$ is odd, positive,
coprime with $B$ and divides $A^2 - 4B = (2n-3)(2n+5)^3$.
If $p$ is an odd prime dividing $B$, then $n \equiv -3 \bmod p$
and so $-D \equiv 1 \bmod p$.
The equation $B x^2 - D y^2 = z^2$ has the solution $(x,y,z)=(1,4,4)$,
so the Hilbert symbol $(B, -D)_p = 1$ for all primes $p$.
We will show:
If $(\xi,\eta) \in E'_n(\mathbb Q)$ with $\xi \neq 0$, then
$(\xi, -D)_p = 1$ for all primes $p$.
Given this, the product formula for the Hilbert symbol implies
$(\xi, -D)_\infty = 1$ and so $\xi > 0$ (since $-D < 0$).
Note that $(\xi, -D)_p = (\xi^2 + A \xi + B, -D)_p$.
We first consider odd $p$. We note that when $\xi$ is not a $p$-adic
integer, then $\xi$ must be a $p$-adic square, so $(\xi, -D)_p = 1$.
So we can assume that $\xi \in {\mathbb Z}_p$. There are three cases.
- $p$ divides neither $B$ nor $D$. If $\xi \in {\mathbb Z}_p^\times$,
then $(\xi, -D)_p = 1$, since both entries are $p$-adic units.
Otherwise, $(\xi, -D)_p = (\xi^2 + A \xi + B, -D)_p = (B, -D)_p = 1$.
- $p$ divides $B$. Then $-D \equiv 1 \bmod p$, so $-D$ is a $p$-adic
square, hence $(\xi, -D)_p = 1$.
- $p$ divides $D$. Then $x^2 + Ax + B \equiv (x + A/2)^2 \bmod p$.
So if $\xi \in {\mathbb Z}_p^\times$, then $\xi$ must be a square
mod $p$, and $(\xi, -D)_p = 1$. If $\xi$ is divisible by $p$,
then as before, $(\xi, -D)_p = (\xi^2 + A \xi + B, -D)_p = (B, -D)_p = 1$.
It remains to consider $p = 2$. If $n \equiv 1 \bmod 4$, then
$-D \equiv 1 \bmod 8$, so $(\xi, -D)_2 = 1$ for all $\xi$.
If $n \equiv 3 \bmod 4$, then $-D \equiv 5 \bmod 8$, so
$(\xi, -D)_2 = (-1)^{v_2(\xi)}$, and we have to show that the
2-adic valuation of $\xi$ must be even. Note that in this
case $v_2(B) = 6$ and $A \equiv -3 \bmod 8$.
If $v_2(\xi)$ is odd, then exactly one of
the three terms $\xi^3$, $A \xi^2$, $B \xi$ has minimal 2-adic
valuation, which must be even, so it cannot be the first or
the third term. This reduces us to $\nu := v_2(\xi) \in \{1,3,5\}$.
One then easily checks that
$\xi(\xi^2 + A\xi + B) = 4^\nu u$ with $u \equiv -1 \bmod 4$
when $\nu = 1$ or $5$ and $u \equiv -3 \bmod 8$ when $\nu = 3$.
In all cases, $u$ cannot be a square, and so points with
$x$-coordinate $\xi$ cannot exist. This concludes the proof.
Note that when $n$ is even, we have $-D \equiv 3 \bmod 4$ and also
$v_2(B) = 5$, so we lose control over the 2-adic Hilbert symbol.
This is the previous version of this answer, which I leave here,
since it may contain some points of interest.
The equation $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = n$
gives rise to the elliptic curve
$$E_n \colon a^3 + b^3 + c^3 + abc - (n-1)(a+b)(b+c)(c+a) = 0.$$
You are asking for rational points on this curve (such that $a+b, b+c, c+a \neq 0$).
For odd positive $n$ up to and including 17, this is a curve
of rank zero (with 6 rational points), whereas for $n = 19$, it has rank 1.
Therefore $E_{19}$ has infinitely many rational points, and your equation
has infinitely many solutions for $n = 19$. I'll do the computations and
find one explicitly.
EDIT: As pointed out by Jeremy Rouse in a comment below, the integral
solutions for $n = 19$ are not positive. More precisely, the real
points $E_n(\mathbb R)$ form two connected components (the discriminant
of $E_n$ is positive), and it is the non-identity component that
contains points with all positive coordinates (taking as the identity
one of the six points like $(1:-1:0)$ or $(1:1:-1)$). So the question
is whether there is odd $n$ such that there is a rational point on
the non-identity component; then the rational points will be dense
on this component and so there will be positive solutions. So far,
no such $n$ turned up, even though there are many such that $E_n$
has positive rank.
FURTHER EDIT: I suspect that there really is no odd $n > 0$
such that $E_n$ has rational points on the non-identity component.
One way of checking this for any given $n$ is to do (half of) a
2-isogeny descent on $E_n$. This produces a number of curves
of the form $C_u \colon y^2 = u x^4 + v x^2 + w$ where $v = 4n(n+3)-3$
and $uw = 32(n+3)$ that are unramified double covers of $E_n$.
We consider the curves $C_u$ that have points over all completions of
$\mathbb Q$. Then every rational point on $E_n$ is the image of a
rational point on one of these curves $C_u$. Doing the computation,
one obtains a set of curves $C_u$ that all have $u > 0$ (this is
only experimental; I checked it for $n$ up to 9999). But if $u > 0$,
then [$C_u$ has only one real component — this is wrong, but
the following is OK] the image of $C_u(\mathbb R)$
in $E_n(\mathbb R)$ is the identity component, so there can be no
rational point on the other component. My feeling is that there might be
a Brauer-Manin obstruction to the existence of rational points on
the non-identity component for odd $n$, but I don't have enough time
to check this. A possible approach would be to note that
$$E'_n \colon y^2 = x \bigl(x^2 + (4n(n+3)-3)x + 32(n+3)\bigr)$$
is isomorphic to $E_n$. If we can find a positive integer $d(n)$
such that for all rational points $(\xi,\eta) \in E'_n(\mathbb Q)$
(with $\xi \neq 0$) the product $\prod_p (\xi, -d(n))_p$ of Hilbert
symbols (over all finite places) is always $+1$, then the claim
follows from the product formula for the Hilbert symbol and
$(\xi, -d(n))_\infty = -1$ for $\xi < 0$.
SUCCESS: For odd $n \ge 3$, $d(n) = 2n-3$ works. One can check that
$(\xi, 3-2n)_p = 1$ for all primes $p$. Details later (it is getting late).
Actually, $d(n) = -5-2n$ works better. See above.
Note that for even $n$, there usually are $C_u$ with $u < 0$ when $E_n$
has positive rank (the first exception seems to be $n = 40$). So I would
expect the Brauer-Manin obstruction to result from an interaction between
$p = 2$ and the infinite place.
For $n = 4$, the curve has also rank 1, which explains the existence
of solutions. I'll try to check if there are smaller ones than that
given by you.
EDIT: The given solution is really the smallest (positive) one. The next larger
one has numbers of 167 to 168 digits.
