Romanian Mathematical Olympiad

We use induction on $n$; the base case $n=1$ is obvious.

Suppose now that the conclusion holds for some positive integer $n$ and consider $m \le (n+1)!$. Then $m = (n+1)q + r$, $q, r \in \mathbb{N}$, $0 \le r \le n$. It is easily noticed that $q \le n!$, therefore $q$ can be written as a sum of (at most) $n$ divisors of $n!$ – denote them $d_1, \dots, d_k$, $k \le n$, whence

$$m = (n + 1)d_1 + (n + 1)d_2 + \dots + (n + 1)d_k + r$$

and $(n+1)d_i$ are divisors of $(n+1)!$.
If $r = 0$, the induction step is finished. Otherwise $r > 0$, $r$ divides $(n+1)!$ and $r < n+1 \le (n+1)d_i$, therefore $m$ can be written as the sum of $k+1$ distinct divisors of $(n+1)!$, with $k+1 \le n+1$ and we are done.