$2n+1$ Subsets whose cardinality is a multiples of 6 Consider $2n+1$ subsets $S_1,S_2,...,S_{2n+1}$ of $\{ 1,2,...,n\}$.
Can anybody show (or provide a counterexample) that if $|S_i|$ is not a multiple of 6 for all $i\in \{ 1,2,...,2n+1\}$, then there are $i\neq j\in \{ 1,2,...,2n+1\}$ such that $|S_i\cap S_j|$ is not a multiple of 6?
I found it on a forum without any answers and I believe there are no counterexample.
Any help or reference would be appreciated.
 A: Let us assume the contrary, i.e., that for some $n$ there exist subsets $S_1,\dots,S_{2n+1}\subset\{1,2,\dots,n\}$ such that $6\nmid |S_i|$ for all $i$, but $6\mid|S_i\cap S_j|$ for all $i\ne j$.
Let $A\subset \{1,2,\dots,2n+1\}$ such that $i\in A$ iff $2\nmid|S_i|$.
Let $B\subset \{1,2,\dots,2n+1\}$ such that $i\in B$ iff $3\nmid|S_i|$.
Clearly, we have $A\cup B = \{1,2,\dots,2n+1\}$, and thus
$$|A|+|B|\geq |A\cup B| = 2n+1.$$
Then we have $|A|>n$ or $|B|>n$ (or both).
Consider the case $m:=|A|>n$ (the other case is considered similarly). Construct a 01-matrix $M$ of size $m\times n$, where the rows are indexed by elements of $A$ such that for all $a\in A$ and $j\in\{1,2,\dots,n\}$, $M_{a,j} = 1$ iff $j\in S_a$.
It follows that $MM^T$ is an $m\times m$ matrix with numbers $|S_a|$ ($a\in A$) on the diagonal, and multiples of $6$ off the diagonal. That is, over the field $\mathbb{F}_2$, $MM^T$ forms the identity matrix and has rank $m$. Then the rank of $M$ over $\mathbb{F}_2$ is at least $m$, but on the other hand it is at most the number of columns $n$, which contradicts $m>n$.
