APMO

Consider four cases: - $n\le 5$. Then $\lfloor \frac{(n-1)!}{n(n+1)}\rfloor =0$ is an even number.

- $n$ and $n+1$ are both composite (in particular, $n\ge 8$). Then $n=ab$ and $n+1=cd$ for $a,b,c,d\ge 2$. Moreover, since $n$ and $n+1$ are coprime, $a,b,c,d$ are all distinct and smaller than $n$, and one can choose $a,b,c,d$ such that exactly one of these four numbers is even. Hence $\frac{(n-1)!}{n(n+1)}$ is an integer. As $n\ge 8>6$, $(n-1)!$ has at least three even factors, so $\frac{(n-1)!}{n(n+1)}$ is an even integer.

- $n\ge 7$ is an odd prime. By Wilson's theorem, $(n-1)!\equiv -1(\mathrm{mod}n)$, that is, $\frac{(n-1)!+1}{n}$ is an integer, as $\frac{(n-1)!+n+1}{n}=\frac{(n-1)!+1}{n}+1$ is. As before, $\frac{(n-1)!}{n+1}$ is an even integer; therefore $\frac{(n-1)!+n+1}{n+1}=\frac{(n-1)!}{n+1}+1$ is an odd integer. Also, $n$ and $n+1$ are coprime and $n$ divides the odd integer $\frac{(n-1)!+n+1}{n+1}$, so $\frac{(n-1)!+n+1}{n(n+1)}$ is also an odd integer. Then $$\lfloor \frac{(n-1)!}{n(n+1)}\rfloor =\frac{(n-1)!+n+1}{n(n+1)}-1$$ is even.

- $n+1\ge 7$ is an odd prime. Again, since $n$ is composite, $\frac{(n-1)!}{n}$ is an even integer, and $\frac{(n-1)!+n}{n}$ is an odd integer. By Wilson's theorem, $n!\equiv -1(\mathrm{mod}n+1)⟺(n-1)!\equiv 1(\mathrm{mod}n+1)$. This means that $n+1$ divides $(n-1)!+n$, and since $n$ and $n+1$ are coprime, $n+1$ also divides $\frac{(n-1)!+n}{n}$. Then $\frac{(n-1)!+n}{n(n+1)}$ is also an odd integer and $$\lfloor \frac{(n-1)!}{n(n+1)}\rfloor =\frac{(n-1)!+n}{n(n+1)}-1$$ is even.