Mongolian Mathematical Olympiad

First we shall prove that a number which can be represented as sum of consecutive natural numbers can not be represented in the form $n=2^k$. $$n=m+(m+1)+(m+2)+\cdots +(m+k)=\frac{(k+1)(2m+k)}{2}$$ 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 $1,2,4,8,16,64,128,256,512,1024$.