Lower bounds for inversion counting in comparison model?
I'm not totally sure, but it seems you can get $\Omega(n \log n)$ lower bound.

Suppose in the worst case we use $o(n \log n)$ queries. Consider a decision tree for our algorithm. Cause of depth of the tree there should be a leaf such that there are at least two permutations which correspond to it.

Now consider all queries with answers on a path from the root to the leaf. We know that they set a partial, not total, order. Consider two elements, which can not be compared. Take any permutation, which corresponds to this order, and swap these elements.

I don't have a strict proof, but it seems to be true, that after such a swap number of inversions should be changed. If so, we broke our algorithm and obtained lower bound.