<s>[INST] 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 $n!$.
 [/INST] People do not seem to be going along with my comment. So this is CW, and directly from the answer by Sanchez.
For $n=2,$ we need only 1 divisor of $2!,$ as $1=1.$
For $n=3,$ we need only 2 divisors of $3!=6,$ as $1=1, 2=2,3=3,4=3+1,5=3+2.$
Induction hypothesis: for some $n \geq 2,$ we need at most $(n-1)$ distinct divisors of $n!$ to write any $1 \leq x < n!$ as a sum.
Induction step (Sanchez, above). Let $N = n+1.$ Let $1 \leq x < N! = (n+1)!$ Write
$$  x = N q + r, \; \; \mbox{with} \; \; 0 \leq r < N.   $$
Because $q < (N-1)! = n!,$ we need at most $(n-1) = (N-2)$ divisors of $n!$ to write $q$ as a sum. So
$$ q = \sum_{i=1}^{n-1} d_i,   $$
where each $d_i | n!$ Therefore each $Nd_i | N!$
At this stage, we have at most $N-2$ divisors of $N!$ What about $r?$ Well, $r < N,$ so it is automatically a factor of $N!$ So we have finished the decomposition as a sum with at most $(N-1)$ divisors of $N!,$ where $N=n+1.$
CONCLUSION: For all $N \geq 2,$ every integer $1 \leq x < N!$ can be written as the sum of (at most) $N-1$ distinct divisors of $N!$
SUGGESTION: try it for $N=4, \; \; N! = 24.$
NEVER MIND, do it myself. Aliquot divisors 1,2,3,4,6,8,12. 
$$1=1,2=2,3=3,4=4,5=4+1,6=4+2,7=4+3, 8=8,9=8+1,10=8+2,  $$
$$11=8+3, 12= 12, 13 = 12+1, 14 = 12+2, 15 = 12+3, 16 = 12+4,$$
$$17=12+4+1,   18=12+6,19=12+4+3,20=12+8,$$
$$21=12+8+1,22=12+8+2,23 = 12+8+3.   $$
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