Answer: $k=2$ is indeed the maximum value. Example: Let the $2021$ points be $a_1, a_2, a_3,..., a_{2021}$. Suppose $a_1, a_3, a_5,..., a_{2021}$ are colored by $1$ and $a_2, a_4, a_6,..., a_{2020}$ are colored by $2$. This clearly works. Now let’s show $k\leq 2$. Suppose the contrary. Let $k\ge 3$. Then, there is a color such that the number of points painted in this color is at most $673$. Let this color be $1$. (Number of the points which are not colored by $1$ is at least $2021-673=1348$) Definition 1: Let's define an arc $Y_i$ for each point $a_i$, which is not colored by $1$, such that this arc is the shortest arc that contains the point $a_i$ and at least half of the points on it are colored by $1$. Claim 1: One of the endpoints of $Y_i$ is the point $a_i$. Proof: Suppose the contrary. Let the endpoints of $Y_i$ be $m$ and $n$. Assume there are $z_1$ points between $m$ and $a_i$ ($m$ included, $a_i$ not) and $z_2$ of them are colored by $1$; $t_1$ points between $n$ and $a_i$ ($n$ included, $a_i$ not) and $t_2$ of them are colored by $1$. We know $z_2<\frac{z_1+1}{2}$ (otherwise the arc starts from the point $m$ and end in $a_i$ would be a suitable arc for the point $a_i$ and it is shorter than $Y_i$) so $z_2\leq \frac{z_1}{2}$. Similarly, $t_2\leq \frac{t_1}{2}$. Thus, $z_2+t_2\leq \frac{z_1+t_1}{2}$. We know $z_2+t_2\ge \frac{z_1+t_1+1}{2}$. Contradiction. Claim 2: Second endpoint of the arc $Y_i$ is colored by $1$. Proof: Suppose the contrary. Let the second endpoint of $Y_i$ be $m$ and assume $m$ is not colored by $1$. Let’s say there are $z_1$ points between $m$ and $a_i$ (both included) and $z_2$ of them are colored by $1$. We know $z_2\ge \frac{z_1}{2}$. Remove the point $m$ from the arc $Y_i$. New arc is shorter than $Y_i$ and at least half of the points on it are colored by $1$ since $z_2>\frac{z_1-1}{2}$ ($m$ is not colored by $1$). Contradiction. Definition 2: Let’s say second endpoint of the arc $Y_i$ be $b_i$. Claim 3: Exactly half of the points on the arc $Y_i$ are colored by $1$. Proof: Suppose the contrary. Let’s say there are $z_1$ points between $a_i$ and $b_i$ (both included) and $z_2$ of them are colored by $1$. Assume $z_2>\frac{z_1}{2}$. Then $z_2\ge \frac{z_1+1}{2}$. Remove the point $b_i$ from the arc $Y_i$. New arc is shorter than $Y_i$ and at least half of the points on it are colored by $1$ since $z_2-1\ge \frac{z_1-1}{2}$. Contradiction. For each arc $Y_i$, consider the point $b_i$. We know all $b_i$ points are colored by $1$ (Claim 2). There is at most $673$ points colored by $1$ and at least $1348$ arcs. By Pigeonhole Principle, we get there exist $3$ arcs which have the same endpoint. WLOG $Y_m,Y_n,Y_p$ have the same endpoint, $b_1$. These $3$ arcs are on the same circle and they all end at same point. So, there should be $2$ arcs one covering the other. WLOG $Y_m$ covers $Y_n$. Let $a_m$ and $a_n$ be their starting point. Let’s say there are $z_1$ points between $a_m$ and $a_n$ ($a_m$ included, $a_n$ not) and $z_2$ of them are colored by $1$; $t_1$ points between $a_n$ and $b_1$ (both included) and $t_2$ of them are colored by $1$. By (Claim 3), we know: $t_2=\frac{t_1}{2}$ (arc $Y_n$) and $z_2+t_2=\frac{z_1+t_1}{2}$ (arc $Y_m$). Thus, $z_2=\frac{z_1}{2}$. Consider the arc starts at the point $a_m$ and ends at point just before $a_n$ (this arc is $Y_m-Y_n$). On this arc, there are $z_1$ points and $z_2$ of them are colored by $1$. Also, $z_2=\frac{z_1}{2}$. So, this arc is suitable for the point $a_m$. But $Y_i$ was the shortest arc which is suitable for $a_m$. Contradiction. In conclusion, $k\leq 2$.