Considering a fixed positive integer $k$ with the given property, by plugging $(m, n) \leftarrow (k, 1)$ and $(m, n) \leftarrow (k^2 + k - 1, k + 1)$ we find that both
$$\frac{k+1}{k^2 - k^2 + 1} = k + 1$$
and
$$\frac{(k^2 + k - 1) + (k + 1)}{(k^2 + k - 1)^2 - k(k^2 + k - 1)(k + 1) + (k + 1)^2} = \frac{k^2 + 2k}{k + 2} = k$$
are not composite. Since they are of different parity, it follows that $k = 1$ or $k = 2$.

1) If $k = 1$, then $\frac{m+n}{m^2-mn+n^2} \le 2$, satisfying the conditions of the problem. Indeed,
$$\frac{m+n}{m^2-mn+n^2} \le 2 \iff (m-n)^2 + m(m-1) + n(n-1) \ge 0.$$

2) $k = 2$ does not satisfy the conditions of the problem. For example, plugging $(m, n) \leftarrow (5, 4)$ into $\frac{m+n}{m^2-2mn+n^2} = \frac{m+n}{(m-n)^2}$ yields composite number 9.

Therefore, $k = 1$ is the only positive integer with the given property.