This problem arose in the study of Latin squares with a large number of subsquares, although it appears interesting in its own right.

> **Question**: What is the maximum cardinality of a family $F \subseteq 2^{[n]}$ of subsets of $[n]:=\{1,2,\ldots,n\}$ for which any two distinct $A,B \in F$ satisfy $|A \cap B| \leq \tfrac{1}{2} \min(|A|,|B|)$?

Some observations:

  - We have the trivial lower bound $\max |F| \geq {n \choose 2}$ by taking all the subsets of size 2.
  - When $n \geq 3$, $F$ should not have any sets of size 1.  If $\{a\} \in F$, then we can replace it by $\{a,x\}$ for all $x \in [n] \setminus \{a\}$ for any $x$ that belongs to a set of size 2 or more (since no other set in $F$ can contain an $a$).  If every set has size 1, then $|F|=n$ which can be beaten.
  - $F$ should not have any sets of size 3.  If $\{a,b,c\} \in F$, then we can replace it by $\{a,b\},\{a,c\},\{b,c\}$ (since any other set in $F$ may intersect $\{a,b,c\}$ in at most one element).
  - With the above simplifications in mind, I wrote a backtracking algorithm which says that for $3 \leq n \leq 7$ (and it's progressing through $n=8$), that $F$ is uniquely maximized when $F$ consists of all 2-subsets.