Consider the function $f$ from the positive integers into themselves. We need to determine all such functions such that for every collection of positive integers $x_1, x_2, \ldots, x_k$, the expression

$$f(x_1)! + f(x_2)! + \cdots + f(x_k)!$$

is divisible by

$$x_1! + x_2! + \cdots + x_k!.$$

Given that $k \geq 2$, let's analyze the problem by initially testing simple cases and then move towards a general solution.

### Step 1: Substituting $x_1 = x_2 = \cdots = x_k = x$

Consider the case where $x_1 = x_2 = \cdots = x_k = x$. The condition reduces to:

$$k \cdot f(x)! \equiv 0 \pmod{k \cdot x!}.$$

This can be simplified to:

$$f(x)! \equiv 0 \pmod{x!}.$$

The most straightforward solution would be the case $f(x) = x$, resulting in $f(x)! = x!$. This satisfies the divisibility condition since $x! \equiv 0 \pmod{x!}$ holds trivially for any $x$.

### Step 2: Generalization and Testing

Now, consider the general function $f$ where $f(n) \neq n$. We need to show whether $f(n) = n$ is uniquely valid or if other functions also satisfy the condition:

Assume $f(n) \neq n$ for some $n$. Consider $x_1 = x_2 = \cdots = x_{k-1} = n$ and $x_k = f(n)$.

For divisibility, we should have:

$$(k-1) \cdot f(n)! + n! \text{ is divisible by } (k-1) \cdot n! + f(n)!.$$

By the properties of factorials, if $f(n) < n$, then $f(n)!$ does not divide $n!$, causing issues in divisibility unless $k \cdot f(n)! \geq n!$. Similarly, if $f(n) > n$, $f(n)!$ exceeds any term of the form $n!$, contradicting divisibility.

### Conclusion

Therefore, the function $f(n) = n$ ensures $f(x_1)! + f(x_2)! + \cdots + f(x_k)!$ is divisible by $x_1! + x_2! + \cdots + x_k!$ for all choices of $x_1, x_2, \ldots, x_k$.

Thus, the only solution is:
$$\boxed{f(n) = n}.$$