351*. Find all natural $n>1$, for which $(n-1)!$ is divisible by $n$.
$\triangle$ Consider two possibilities.

1) Let the number $n-$ be prime.

But then the number ( $n-1$ )! is coprime with $n$ and, consequently, cannot be divisible by $n$. Therefore, this case is ruled out.

2) Let the number $n-$ be composite

If in this case $n$ can be factored into two distinct factors $a$ and $b$, both greater than 1, then the numbers $a$ and $b$ are included in the factorization

$$
(n-1)!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot(n-1)
$$

at different positions. Hence, ( $n-1$ )! is divisible by $a b=n$. This means that all such composite $n$ satisfy the condition of the problem.

Now let $n=a^{2}(a>1)$. For $a=2$, i.e., for $n=4$, we get: 3 ! is divisible by 4, which is false. If $a>2$, then for the product

$$
1 \cdot 2 \cdot 3 \cdot \ldots \cdot\left(a^{2}-1\right)
$$

to be divisible by $n=a^{2}$, it is sufficient that the factors include the numbers $a$ and $2 a$, i.e., that

$$
a^{2}-1 \geqslant 2 a, \quad a^{2}-2 a-1 \geqslant 0
$$

Then $a \geqslant 1+\sqrt{2}$. But this inequality holds for all $a>2$. Therefore, all $n=a^{2}$, where $a>2$, also work.

Answer: all composite numbers, except 4.