Is it possible to find an example of a family $\mathcal{F}$ of $n$ finite distinct non-empty sets, a universe of maximum size $n/4$, with at least $\lfloor \frac{2}{3}{n \choose 2} \rfloor$ unordered couples of sets with at least one element in common between the two sets, and no element belonging to at least $n/2$ sets of the family?



Existence of a family of sets with some properties
Take a finite projective plane: for the sake of concreteness, the Fano plane $PG(2, 2)$. It has seven points $P$ and seven lines $L$, where each line goes through three points and each pair of lines intersects.


Take a second set $S$ which is disjoint from $P$ and an element $x$ which is not in either. Consider $$\mathcal{F} = L \times 2^S \cup \{\{x\}\}$$ and let $n = |\mathcal{F}| = 7 \cdot 2^{|S|} + 1$. Every element of $P$ is in $\frac{3}{7}(n-1)$ sets; every element of $S$ is in $\frac{1}{2}(n-1)$ sets; $x$ is in one set. Every pair of sets which does not include $\{x\}$ has a non-empty intersection. The size of the universe is $|S| + 8$, so this family satisfies all of the constraints provided that $4|S| + 32 \le 7 \cdot 2^{|S|} + 1$, which is true if $|S| \ge 3$.




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For more general values of $n$, let $a$ be the largest integer such that $7 \cdot 2^a < n$. If we take two disjoint sets $S\_1, S\_2$ with $|S\_1| = |S\_2| = a$ then $L \times \left(2^{S\_1} \cup 2^{S\_2} \right)$ has more than $n$ sets, but each element of $S\_1 \cup S\_2$ occurs exactly $7 \cdot 2^{a-1} < \frac n2$ times. Each element of $P$ occurs exactly $3 \cdot 2^{a+1}$ times, so when choosing $\mathcal{F} \subset L \times \left(2^{S\_1} \cup 2^{S\_2} \right)$ we should be careful to keep the projection onto $L$ balanced, but if we do this then each element of $P$ will occur no more than $\frac{3}{7}n + 3$ times.


Then we simply require that the universe be sufficiently small. Now the size of the universe is $7 + 2a$, and we find that the construction works for $n \ge 44$.


