Problem 1697

$a_1, a_2, ..., a_n$ Without the loss of generality, let us assume that $a_1 < a_2 < ... < a_n$. Then $(a_i + 1)! \le a_i!(a_n + 1)$, and equality is reached only when $i = n$.

Let $(a_1 + 1)! + (a_2 + 1)! + ... + (a_n + 1)! = N(a_1! + a_2! + ... + a_n!)$ for some natural number $N$. Then the condition above yields $N \le a_n + 1$, moreover, if $n > 1$, then $N < a_n + 1$.

Let now $n > 1$, then we have that $N \le a_n$. Let us prove that it is impossible by showing that:
$$(a_1+1)!+(a_2+1)!+...+(a_n+1)! > a_n(a_1!+a_2!+...+a_n!)$$

Let's consider the following expression:
$$(a_1+1)!+(a_2+1)!+...+(a_n+1)! - a_n(a_1!+a_2!+...+a_n!) = \sum_{i=1}^{n} a_i!(a_i+1-a_n).$$
The last term is equal to $a_n!$. If $a_{n-1} = a_n - 1$, then the penultimate term is zero. It is enough to consider the following:
$$\begin{aligned} \sum_{i=1}^{n} a_i!(a_i+1-a_n) &\le \sum_{i=1}^{a_n-1} a_i!(a_n-1-i) = \sum_{i=1}^{a_n-2} a_i!(a_n-1-i) \ &\le \sum_{i=1}^{a_n-2} i!a_n = a_n \sum_{i=1}^{a_n-2} i! \le \\ a_n(a_n-2)(a_n-2)! < a_n! \Rightarrow \\ (a_1+1)!+(a_2+1)!+...+(a_n+1)! - a_n(a_1!+a_2!+...+a_n!) > 0, \end{aligned}$$
which yields the desired result. Hence, there are no such sets for $n > 1$.