Let $L_n=\text{lcm}[1,2,\dots,n]$. First, we prove by induction that for each positive integer $n$, we can find some $A_n$ such that given a list of numbers in $\{1,2,\dots,n\}$ whose sum is $\ge A_n$, we can select a number of them whose sum is $L_n$. The induction proof. Either there are enough $n$'s in our list which add up to $L_n$ and we're done, or the numbers in the list from $\{1,2,\dots,n-1\}$ have sum $\ge A_n-L_n+n$. In the latter case, we can keep selecting groups of them which add up to $L_{n-1}$ while the sum of the remaining numbers is at least $A_{n-1}$, which can be blocked together into a block with sum $L_n$. So we just need $A_n-L_n+n-L_n+L_{n-1}\ge A_{n-1}$, so we define $A_1=1$ and \[ A_n:=A_{n-1}+(2L_n-L_{n-1}-n).

To solve the problem, just choose $n\ge \max S$ and choose $N$ to be a multiple of $L_n$ for which $N\ge A_n$.

Problem wrote: A list of integers from $\{1,2,\dots,n\}$ adds up to $kn!$. Prove that the list can be partitioned into $k$ groups who add up to $n!$. Solution. The claim is obvious for $n=1$; we induct now from $n-1$ to $n$. Since some two of the sums \[ 0,x_1,x_1+x_2,\dots,(x_1+\dots+x_n) \]have the same residue mod $n$, taking their difference shows that from any $n$ integers, we can select some whose sum is divisible by $n$. We can thus repeat the process of removing collections of $\le n$ numbers in the list whose sum is divisible by $n$, until we have $\le n$ numbers left, which also form such a collection. If we take care to begin the process by forming one-member collections out of all the numbers $n$, each collection will have sum $<n^2$, and thus the sums are $na_1,na_2,\dots,na_\ell$ where $a_i\in\{1,2,\dots,n-1\}$ for each $i$. Applying the induction hypothesis to the $a_i$'s completes the proof. $\blacksquare$