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,
$$n-1 = 2^{m_1} + \dots + 2^{m_k},$$
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 $2^n - 1 = 2^{n-1} + 2^{n-2} + \dots + 2^1 + 2^0$, we obtain:
$$n(2^n - 1) = (2^{m_1} + \dots + 2^{m_k})2^n + [(2^{n-1} + \dots + 2^1 + 2^0) - (2^{m_1} + \dots + 2^{m_k})] = \\ = \underbrace{2^{n+m_1} + \dots + 2^{n+m_k}}_{A} + \underbrace{[(2^{n-1} + 2^{n-2} + \dots + 2^1 + 2^0) - (2^{m_1} + \dots + 2^{m_k})]}_{B}.$$
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 $2$.