VIII OBM

The sum of $k$ consecutive integers is $(m+1)+(m+2)+\cdots +(m+k)=\frac{k(2m+k+1)}{2}$. So we require $k(2m+k+1)=2n$. Note that $k$ and $2m+k+1$ have opposite parity and that $k<2m+k+1$.

Now suppose $2n=ab$ with $a$ odd. Then $b$ must be even. So $a$ and $b$ cannot be equal. Take $k$ to be the smaller, then put $m=\frac{|a-b-1|}{2}$ and we have a solution. So the total number of solutions is just the number of odd factors of $2n$.