Find a sequence of $n$ integers with $\Omega(n^2)$ inversions
Sorta answered my own question in the comments, so I decided to put this here:

If we take any sequence of $n$ integers, in reverse order like $\{n,n-1,\dots, 2, 1\}$. Then each digit $S[i]$ will have $(i-1)$ inversions, and there are $n$ digits.

Thus $I(n) = \displaystyle\sum_{0}^{n-1} n = 0 + 1 + 2 + \dots + (n-1)$. So the total number of inversions will be $I(n) = \dfrac{n(n-1)}{2}$.

Then see, since $n^2 = O\left(\dfrac{n(n-1)}{2}\right)$, $\dfrac{n(n-1)}{2} = \Omega(n^2) $.

Thus $I(n) \in \Omega(n^2)$, and we have a sequence of $n$ digits with $\Omega(n^2)$ inversions.