Baltic Way 2019

Answer: $k=1010=\lfloor n/2\rfloor$, where $n$ is the number of points.

We say that discs separates two points if it contains only one of them.

Estimation. Consider a set of $n$ points on a circle. Then any disc can separate at most two consecutive pairs, therefore we need at least $\lfloor n/2\rfloor$ discs.

Example. Let $S_1$ be a set of $n$ points, and $H_1$ be the convex hull of $S$. Let $D_1$ be a disc that contains exactly $\lfloor n/2\rfloor$ points. There are points $A_1,B_1\in S_1$ in the boundary of $H_1$, such that $A_1\in D_1,B_1\notin D_1$ and the (open) interval $A_1{B}_1$ does not contain other points of $S_1$. Let $D_2$ be a disc that separates points $A_1,B_1$ from the other points of $S_1$.

Denote the set $S_1\setminus \{A_1,B_1\}$ by $S_2$ and its convex hull by $H_2$. There are points $A_2,B_2\in S_2$ in the boundary of $H_2$, such that $A_2\in D_2,B_2\notin D_2$ and the (open) interval $A_2{B}_2$ does not contain other points of $S_2$. Let $D_3$ be a disc that separates points $A_2,B_2$ from the other points of $S_2$.

By continuing this procedure we get a set of $\lfloor n/2\rfloor$ discs. (The description of the final step of the process we left to the reader. The last point that has no pair is $A_{\lfloor n/2\rfloor }$.) This set of discs has the desired separating property. Indeed, let $P,Q\in S_1$. If $P=A_i,Q=B_j$ for some $i,j$, then $D_1$ separates them. If $P=A_i$ and $Q=A_j$, $i<j$, (or if $P=B_i$ and $Q=B_j$, $i<j$), then $D_{i+1}$ separates them.