Prove that for any $x \in \mathbb N$ such that $xProve that every positive integer $x$ with $x<n!$ is the sum of at most $n$ distinct divisors of $n!$.

Not quite there, but a start:  As suggested in Wikipedia on practical numbers we will use the greedy algorithm.  First pull out $n!/2$ if that is possible, then $n!/3$, then $n!/4$ and so on, stopping when the remainder is less than or equal to $n$ and skipping denominators that don't divide $n!$.  If $n$ and $x$ are very large, the denominators we use will follow Sylvester's sequence:  $2, 3, 7, 43, 1807, 3263443, 10650056950807,\ldots$ which is given by $a(0)=2, a(n+1)=a(n)^2-a(n)+1$.  To use induction, we need to find a sequence of $m$ denominators that reduce $n!-1$ to something less than $n$.  For $n$ in the range $5-6$ we can use $2,3,8,30$.  For $7$ we can use $2,3,7,45$, for $8-10$ we can use $2,3,7,45,640$.  Then $44$ becomes available at $11$.  It "obviously" works, but I can't prove it.
