Given any possible partial ordering of size $n$, what is the maximum number of comparisons needed to define the partial ordering? As an example, the partial orderings for size 4 are:
1 : No comparisons
2 : $A < B$
3 : $A < B, A < C$
4 : $A < C, B < C$
5 : $A < B, C < D$
6 : $A < B, B < C$
7 : $A < C, B < C, B < D$
8 : $A < B, A < C, A < D$
9 : $A < D, B < D, C < D$
10: $A < C, A < D, B < C, B < D$
11: $A < B, B < C, A < D$
12: $A < D, B < C, C < D$
13: $A < B, B < C, B < D$
14: $A < C, B < C, C < D$
15: $A < B, A < C, B < D, C < D$
16: $A < B, B < C, C < D$
Here, partial orders 10 and 15 each require 4 comparisons in order to be defined. So for 4 elements, at most 4 comparisons are needed to define any partial ordering. I'm wondering if it is known the maximum number of comparisons needed to define any partial ordering of a $n$-element set.
I used this image of the Hasse diagrams of order 4 to construct the list. This may be less confusing to look at than the list I provided. Comparisons here are represented by the number of edges in the graph. https://i.stack.imgur.com/yDLF7.jpg
In other words, you are asking for the maximum number of edges in the Hasse diagram of a partially ordered set with $n$ elements. The answer is $\left\lfloor\frac{n^2}4\right\rfloor$,which by Mantel's theorem is the maximum number of edges in a triangle-free graph on $n$ vertices. That's because the Hasse diagram of a partially ordered set is a triangle-free graph, and because the maximum in Mantel's theorem is attained by the complete bipartite graph $K_{\left\lfloor\frac n2\right\rfloor,\left\lceil\frac n2\right\rceil}$, which can be realized as the Hasse diagram of a partially ordered set of height two.