So right off the bat, this is in fact a homework question. I've been banging my head on this problem for hours. From my naive understanding of inversions, I feel that this is not possible.

Considering for $n=1$ elements, there are no inversions, it already seems like this won't work, but maybe it's just for $n > 1$, so we try $n = 2$. And the best I can seem to get is $1$ inversion, where the set is $\{2,1\}$, thus there is one inversion, since $0 < 1$ but $S[0] > S[1]$. The best I can seem to get is $\{n, n-1, \dots, 2, 1\}$, where there are $n(n-1) - 1$ inversions.

Am I missing something in my approach?