If $X$ is a set and $A,B\subseteq X$ we let the *Hamming distance* of $A,B$ be defined as  $d_H(A,B) = \big|(A\setminus B)\cup(B\setminus A)\big|$. If $\newcommand{\S}{{\cal S}}\S\subseteq {\cal P}(X)$, let the *discrepancy* of ${\cal S}$ be definied by $\newcommand{\d}{\text{discr}}\d(\S)=\min\{d_H(A,B): A\neq B\in\S\}$.

For $n\in\mathbb{N}$, let $[n]=\{1,\ldots,n\}$. Let $X$ be a set and $n\in\mathbb{N}$. Then $[X]^n$ denotes the collection of subsets of $X$ having $n$ elements. 

For $n\in\mathbb{N}$, let $M_n:= \max\{\d(\S):\S\in \big[{\cal P}([n])\big]^n\}$.

Experiments by [Rob Pratt](https://mathoverflow.net/users/141766/robpratt) and [Claude Chaunier](https://mathoverflow.net/users/127616/claude-chaunier) as laid out in the comments below (cheers to Rob and Claude!) seem to point to the following hypothesis: 

> For all $k\in\mathbb{N}$ with $k\geq 2$ we have
> - $M_{2k} = k$ , and
> - $M_{2k+1} = k$ for $k$ even, and $M_{2k+1} = k+1$ for $k$ odd.

**Question.** Is the above hypothesis true?