First we shall prove that a number which can be represented as sum of consecutive natural numbers can not be represented in the form $n = 2^k$.
$$n = m + (m + 1) + (m + 2) + \dots + (m + k) = \frac{(k + 1)(2m + k)}{2}$$
Note that the numbers $k+1$ and $2m+k$ are different by (mod 2). Hence one of these numbers is odd. Now let's prove that any number of the form $n \ne 2^k$ can be represented as sum of consecutive natural numbers.

Thus, $n = 2^h \cdot l$, $l > 1$ is odd. If $2^{h+1}$ then it is sufficient to take $k = 2^{h+1} - 1$ and $m = \frac{l-k}{2} = \frac{l+1-2^{h+1}}{2} = \frac{l+1-2^h}{2}$.

If $2^{h+1} < l$ then setting $k = l - 1$ and $m = \frac{2^{h+1}-k}{2} = \frac{2^{h+1}-l+1}{2}$ we have done. Finally, we concluded desired numbers are $1, 2, 4, 8, 16, 64, 128, 256, 512, 1024$.