We have a set $S$ with $k$ elements, a positive integer $n$, and subsets $S_1, S_2, \cdots, S_n,$ each with n elements. For any two elements $a,b$ of $S$, there are at most two sets $S_i$ containing both $a$ and $b.$ Must k be $\Omega(n^2)?$

If we replace the word "two" with the word "one", we have $k \ge \frac{(n+1)n}{2},$ because $S_i$ contains at least $n+1-i$ elements that are not in $S_1, S_2, \cdots, S_{i-1}.$ In this case, a special case of this is that if we have $n$ pairwise disjoint lines in the plane and a set of points such that each line contains at least $n$ points, we have at least $\frac{n(n+1)}{2}$ points.

One motivation for this question is the generalization of Szemerédi-Trotter to a family of curves satisfying 1) two curves intersect in at most m points and 2) for every two points, at most n curves go through both of those points.

I'm trying to solve a graph theoretical problem assuming only an analogue of condition 2, and this is the easiest case of it. I expect that an extra condition is necessary, but I cannot find any obvious counterexamples.