, 2021

Answer: Yes.

Let us first prove a lemma that if $4$ points $A$, $B$, $C$, $D$ all lie on circle $\Gamma$ and some two points $X$, $Y$ do not lie on $\Gamma$, then these $6$ points are pairs of intersections of three circles, circle $\Gamma$ and two other circles. Indeed, according to the problem statement there are $4$ points among $A$, $B$, $C$, $X$, $Y$ which are concyclic. These $4$ points must include points $X$ and $Y$ because if one of them is not, then the other one must lie on $\Gamma$. Without loss of generality, take $A$, $B$, $X$, $Y$ to lie on the same circle. Similarly for points $A$, $C$, $D$, $X$, $Y$ there must be $4$ points which are concyclic. Analogously, they must include points $X$ and $Y$. Point $A$ cannot be one of them because two circles cannot have more than two common points. Therefore, points $C$, $D$, $X$, $Y$ are concyclic which proves the lemma.

Let us first solve the problem for the case for which there exist $5$ points which lie on one circle $\Gamma$. Label these points $A$, $B$, $C$, $D$, $E$. Let us assume that there exists two points which do not lie on $\Gamma$, label them $X$ and $Y$. According to the previously proven lemma, points $A$, $B$, $C$, $D$, $X$, $Y$ must be the pairwise intersections of $3$ circles. Without loss of generality, let the intersections of $\Gamma$ with one of the other circles be $A$ and $B$ and with the other circle $C$ and $D$. Similarly, $A$, $B$, $C$, $E$, $X$, $Y$ must be the pairwise intersections of three circles one of which is $\Gamma$. This is not possible as none of the points $A$, $B$, $C$ lies on the circumcircle of triangle $EXY$. This contradiction shows that at most $1$ point can lie outside circle $\Gamma$, i.e. at least $2020$ points lie on circle $\Gamma$.

It remains to look at the case for which no $5$ points lie on the same circle. Let $A$, $B$, $C$, $D$, $E$ be arbitrary $5$ points. Without loss of generality, let $A$, $B$, $C$, $D$ be concyclic and $E$ a point not on this circle. According to the lemma, for every other point $F$ and points $A$, $B$, $C$, $D$, $E$, the $6$ points are the intersections of circle $\Gamma$ and some two other circles. But in total, there are $3$ such points because one of the two circles must go through $E$ and some $2$ points out of $A$, $B$, $C$, $D$, while the other circle must go through point $E$ and the other two points out of $A$, $B$, $C$, $D$. There are only three partitions of $A$, $B$, $C$, $D$ into two sets. There is a contradiction, as there are $2021 > 5 + 3$ points in total. $\square$