Saudi Arabia Mathematical Competitions

We claim that those integers are $n=2$ and $n=3$.

Suppose $n$ is a composite integer, $n=ab$, $1<a\le b$. If $a<b$, then $n=ab$ divides $b!$ and $(b+1)!$. Since $b<b+1<n-1$, it follows that $b!$ and $(b+1)!$ yield equal remainders (both $0$) at division by $n$. If $2<a=b$ then $(2a)!$ and $(2a+1)!$ are divisible by $n=a^2$. Since $2a+1<a^2-1=n-1$, it follows that $n$ does not satisfy the desired property.

It remains to settle the case of prime $n$. The value $n=2$ checks; let then $n\ge 3$. From Wilson's theorem we have $$(n-1)!\equiv -1(\mathrm{mod}n)$$ and then $$(n-2)!\equiv 1(\mathrm{mod}n).$$ Since $1!=1$ yields remainder $1$ at division by $n$, it follows $n-2=1$, that is $n=3$, which checks.