Croatian Mathematical Olympiad

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, $$\frac{m+n}{m^2-mn+n^2}\le 2⟺(m-n{)}^2+m(m-1)+n(n-1)\ge 0.$$

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.