49th Mathematical Olympiad in Ukraine

Obviously, $n$ and $(n+1)$ are coprime numbers. Consequently, for this equation we have that $m$ is divisible by $(n+1)$ and $2009$ is divisible by $n^2$. Since $2009=7^2\cdot 41$, this can be achieved in two ways: $n=1$ or $n=7$.

If $n=1$, then $m=2009\cdot 2=4018$.

If $n=7$, then $49m=2009\cdot 8$, therefore $m=41\cdot 8=328$.

So this equation has two solutions.