Let $A$ be a set of $2k$ points in $\mathbb{R}^n$ such that no open set in $\mathbb{R}^n$ of diameter $2$ contains more than $k$ of these points. What is the largest possible distance $r\_n>0$ one can guarantee so that for all $k$ we can pair up $2k$ points under the mentioned condition so that each pair of points is at least $r\_n$ apart? It is not difficult to show that $1\leq r\_n \leq 2$ ($2$ comes from just taking $|A|=2$ and proving that one can always have them at distance $1$, one can build a distance graph so that two point from $A$ are joined if they are at least $1$ apart - such graph has minimum degree at least $k$ and so has a Hamilton cycle). 



Separating pairs of points in R^n
Consider a regular $2k-1$-gon of diameter $2$, meaning the distance between the two most distant vertices is $2$. Then only $k-1$ vertices can be in any set of diameter less than two, becsuse any set of $k$ vertices contains a maximum-distance pair. So the set of vertices plus the center form a set satisfying your condition. As $k$ goes to $\infty$, the polygon approaches a circle of radius $1$, so $r\_n=1$ for $n\geq 2$.


