All symbols in the sequel are denoting integer numbers. Let $n = 2^a b$, $a \ge 0$, $b \ge 1$, $b$ odd. We want to have $n = (m+1) + (m+2) + \dots + (m+k)$, with $m \ge 0$ and $k \ge 2$, hence $k(2m+k+1) = 2^{a+1}b$.

If $b=1$, it follows $k=2^\alpha$, with $1 \le \alpha \le a+1$, but then $2m+k+1 > 1$ is odd, so there are no solutions. Now, for $b > 1$, we can exhibit the required writing.

If $b \ge 2^{a+1} + 1$, then take $m = \frac{1}{2}(b - 2^{a+1} - 1)$ and $k = 2^{a+1}$.

If $b \le 2^{a+1} - 1$, then take $m = \frac{1}{2}(2^{a+1} - 1 - b)$ and $k = b$.