Final Round of National Olympiad

The sum of the numbers on the board is
$$s_n = 1 \cdot n + 2 \cdot (n-1) + 4 \cdot (n-2) + \dots + 2^{n-1} \cdot 1.$$
Let us also define
$$r_n = \frac{s_n}{2^n} = \frac{1}{2} \cdot 1 + \frac{1}{4} \cdot 2 + \frac{1}{8} \cdot 3 + \dots + \frac{1}{2^n} \cdot n.$$

$\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{n+1}} < \frac{1}{2} + \frac{1}{4} + \dots = 1$. Thus $r_n < 2$ always implies $r_{n+1} < \frac{r_n}{2} + 1 < 1 + 1 = 2$. As $r_1 = \frac{1}{2} < 2$, we have $r_n < 2$ for every $n \ge 1$. From there we get that $s_n = 2^n \cdot r_n < 2^{n+1}$ for every $n \ge 1$.