BMO Short List

The identity is the only function satisfying the condition in the statement. Begin by letting the $x$'s be all equal to $n$ to infer that $f(n)!$ is divisible by $n!$, so $f(n)\ge n$ for all positive integers $n$.

Claim. $f(p-1)=p-1$ for all but finitely many primes $p$.

Assume the Claim for the moment to proceed as follows: Fix any positive integer $n$, and let $p$ be a large enough prime, e.g., $p>f(n)!-n!$. Then let one of the $x$'s be equal to $n$ and the remaining $k-1$ be all equal to $p-1$, and use the Claim to infer that the number $$(f(n)!-n!)+(n!+(k-1)(p-1)!)=f(n)!+(k-1)f(p-1)!$$ is divisible by $n!+(k-1)(p-1)!$, and hence so is $f(n)!-n!$. Since $p$ is large enough, this forces $f(n)!=n!$, and since $f(n)\ge n$, it follows that $f(n)=n$, as desired.

Proof of the Claim. If $k$ is even, let $p>f(1)$, and let half of the $x$'s be all equal to $1$ and the other half be all equal to $p-1$, to infer that $f(p-1)!+f(1)!$ is divisible by $(p-1)!+1$. By Wilson's theorem, the latter is divisible by $p$, and hence so is the former. Since $p>f(1)$, the number $f(1)!$ is not divisible by $p$, so $f(p-1)!$ is not divisible by $p$ either, forcing $f(p-1)\le p-1$. Recall now that $f(p-1)\ge p-1$, to conclude that $f(p-1)=p-1$.

If $k$ is odd, let $p>f(2)+\frac{1}{2}(k-3)f(1)$, let $\frac{1}{2}(k+1)$ of the $x$'s be all equal to $p-1$, let one of the $x$'s be equal to $2$, and let the remaining ones (if any) be all equal to $1$, to infer that $\frac{1}{2}(k+1)f(p-1)+f(2)+\frac{1}{2}(k-3)f(1)$ is divisible by $\frac{1}{2}(k+1)(p-1)!+2+\frac{1}{2}(k-3)=\frac{1}{2}(k+1)((p-1)!+1)$. By Wilson's theorem, the latter is divisible by $p$, and hence so is the former. Since $p>f(2)+\frac{1}{2}(k-3)f(1)$, the number $f(2)+\frac{1}{2}(k-3)f(1)$ is not divisible by $p$, so $\frac{1}{2}(k+1)f(p-1)$ is not divisible by $p$ either, forcing $f(p-1)\le p-1$. Recall again that $f(p-1)\ge p-1$, to conclude that $f(p-1)=p-1$.