Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 109 / 4 / iv / Solution

The answer is $\Theta (n^2)$. Take a Steiner triple system, or a partial one of quadratic size, on $\{2,\dots ,n\}$ and adjoin the fixed point $1$ to every triple. Distinct triples meet in zero or one point, so the resulting four-sets meet in one or two points. This gives $\Omega (n^2)$ members. The Ray-Chaudhuri–Wilson theorem with the two allowed intersection sizes gives $O(n^2)$.