For the case of a factorial dividing another factorial, one has that $m! \mid n!$ if and only if $m \le n$, for any two positive integers $m$ and $n$ (yes, they must be strictly positive, because otherwise, $1! \mid 0!$ would be a counterexample).
Also, for the case of a factorial dividing the sum of two factorials, one has that for any three positive integers $m$, $n_1$, and $n_2$, $m! \mid n_1! + n_2!$ if and only if $m \le n_1$ and $m \le n_2$ or $(m, n_1, n_2) = (2, 1, 1)$. In other words, $2!$ is the only factorial dividing the sum of two smaller factorials.
More generally, given any positive integer $k$ and any $k+1$ positive integers $m$, $n_1$, $n_2$, ..., and $n_k$, is it the case that $m! \mid n_1! + n_2! + ... + n_k!$ if and only if $m \le n_j$ for $j=1, 2, ..., k$, with finitely many single exceptions (where $m > n_j$ for $j=1, 2, ..., k$) and finitely many infinite families of exceptions (using the exceptions for the cases of smaller $k$)?
An example of an infinite family of exceptions occurs by observing that $2! \mid 1! + 1! + n!$ for any $n \ge 2$. But actually, this family for the case $k=3$ just uses the exception for the case $k=2$.