Estonija, 2010

For each $n \ge 2$ denote $S_n = a_1 + \dots + a_{n-1}$. Then $a_n = S_n \cdot n$ and for all $n > 2$ we have $S_n = S_{n-1} + a_{n-1} = S_{n-1} + S_{n-1} \cdot (n-1) = S_{n-1} \cdot n$. Hence $S_n = S_{n-1} \cdot n = S_{n-2} \cdot (n-1)n = \dots = S_2 \cdot 3 \cdot \dots \cdot n = \frac{n!}{2}$ because $S_2 = 1 = \frac{1 \cdot 2}{2}$. Consequently $a_n = S_n \cdot n = n! \cdot \frac{n}{2}$ for all $n \ge 2$. Therefore, for $n \ge 2$, $a_n$ is divisible by $n!$ iff $n$ is even, and $n=1$ also satisfies the condition.