For $(\Rightarrow)$ See here http://www.artofproblemsolving.com/community/q1h1201288p5907333 For $(\Leftarrow)$ Let $2n=2^{l+1}t$ for odd number $t \neq 1$ and $l \in \mathbb{Z}^+_0$ If $t > 2^{l+1}$, we have $n=(\frac{t-2^{l+1}-1}{2}+1)+(\frac{t-2^{l+1}-1}{2}+2)+...+(\frac{t-2^{l+1}-1}{2}+2^{l+1})$ If $t < 2^{l+1}$, we have $n=(\frac{2^{l+1}-t-1}{2}+1)+(\frac{2^{l+1}-t-1}{2}+2)+...+(\frac{2^{l+1}-t-1}{2}+t)$

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