1. A building has 4 elevators, each of which can stop at three floors (not necessarily consecutive floors, and not necessarily the lowest floor). For any two floors in the building, there is at least one elevator that can stop at both. How many floors can this building have at most?
2. Let the building have $n$ floors, then the number of floor pairs is $\frac{n(n-1)}{2}$. Each elevator stops at 3 floors, which gives $\frac{3 \times 2}{2}=3$ floor pairs, so, $4 \times 3 \geqslant$ $\frac{n(n-1)}{2}$. Therefore, $n \leqslant 5$.

When $n=5$, the floors served by the four elevators are $(1,4,5),(2,4,5),(3,4,5),(1,2,3)$