Trying out small $m$ it is easy to see that $n=m+1$ is the solution when $m$ is odd. Let $S^-$ be the sum of the non-positive numbers and let $|S^-|=k$ and $S^+$ be the sum of the non-negative numbers. Also WLOG $S^-+S^+ \geq 0$. The equation becomes: $-2S^- = m \leq 2k$. Similarly looking at $S^+$ we have $S^+ \leq n-k$ but from the WLOG we have $-S^- \leq n-k$. From here we get $\frac{m}{2} \leq k$ and $\frac{m}{2} \leq n-k$ by adding and differentiating the cases when $m$ is odd or even we get the needed bounds. Now a construction isn't hard at all.