Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

We shall prove that any such configuration contains at most $n$ chords and the upper bound is achieved, so the required maximum is $n$.

To this end, fix an orientation of the circle and relabel the points $1$ through $n$ in the corresponding circular order. Consider a configuration of chords $[ij]$, $i \neq j$, with pairwise non-empty intersections. Assign to each point $i$, which is an endpoint of at least one chord, the first point $i'$ following $i$, to which it is connected. We now show that by deleting the chords $[ii']$, no chord is left, so the number of chords in the configuration does not exceed $n$.

Suppose, if possible, that some chord $[ij]$ is left. Then $i, i', j, j'$ are in circular order around the circle, so the chords $[ii']$ and $[jj']$ do not meet – a contradiction.

A maximal configuration is given by the $n$ chords $[1i]$, $i = 2, 3, \dots, n$, and $[2n]$.

(The $n$ points could be located anywhere in the plane, or the chords could be Jordan arcs; the topic is related to John Conway's thrackle conjecture.)