The sum of $k$ consecutive integers is $(m+1) + (m+2) + \dots + (m+k) = \frac{k(2m+k+1)}{2}$. 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 $2n$.