Nice question.

First off you can see that the maximum number of floors is less than 12 because then that would mean that all floors (if they have a lift at all) can only be served by 1 lift (4x3=12).

For each floor to have at least 2 lifts stopping there, there can in fact be no more than 6 floors (4x3/2). Now the job is to prove that 6 floors can be served by these 4 lifts.

So try it out, by trial and error:

a: 1,2,3

b: 1,4,5

c: 2,3,6

d: 4,5,6

yep, that works. Each of the 4 lifts (a, b, c, d) are stopping at 3 floors, and each floor has exactly 2 lifts stopping at each.

So the maximum is 6.

This question can I believe be solved rigirously using techniques from graph theory, but this question is simple enough not to have to get those big guns out.