XXVII Olimpiada Matemática Rioplatense

First, we observe that $n<2^n$ for every positive integer $n$ (it can be proved easily by induction). Consider a positive integer $n\ge 2$ (for $n=1$ we have $1\cdot (2^1-1)=2^0$). Let $k$ be the number of ones in the binary representation of $n-1$. Then, $$n-1=2^{m_1}+\cdots +2^{m_k},$$ where the exponents $m_1,\dots ,m_k$ are all distinct and smaller than $n$, due to our initial remark. We express $n(2^n-1)$ as follows: $$n(2^n-1)=(n-1)2^n+[(2^n-1)-(n-1)],$$ and, since $2^n-1=2^{n-1}+2^{n-2}+\cdots +2^1+2^0$, we obtain: $$\begin{gathered} n(2^n-1)=(2^{m_1}+\cdots +2^{m_k})2^n+[(2^{n-1}+\cdots +2^1+2^0)-(2^{m_1}+\cdots +2^{m_k})]= \\ =\underset{A}{\underbrace{2^{n+m_1}+\cdots +2^{n+m_k}}}+\underset{B}{\underbrace{[(2^{n-1}+2^{n-2}+\cdots +2^1+2^0)-(2^{m_1}+\cdots +2^{m_k})]}}. \end{gathered}$$ Note that $A$ is a sum of $k$ distinct powers of $2$ whose exponents are all greater than or equal to $n$. After cancelation of terms (recall that $m_1,\dots ,m_k$ are smaller than $n$), it turns out that $B$ is a sum of $n-k$ distinct powers of $2$ whose exponents are all smaller than $n$. Therefore, the expression obtained is a sum of $k+(n-k)=n$ distinct powers of $2$.