$2021$ points are given on a circle. Each point is colored by one of the $1,2, \cdots ,k$ colors. For all points and colors $1\leq r \leq k$, there exist an arc such that at least half of the points on it are colored with $r$. Find the maximum possible value of $k$.
1. **Initial Setup and Example:**
   - We are given 2021 points on a circle, each colored with one of \( k \) colors.
   - We need to find the maximum possible value of \( k \) such that for every color \( 1 \leq r \leq k \), there exists an arc where at least half of the points on it are colored with \( r \).
   - Example: If \( k = 2 \), we can color the points alternately with two colors, say 1 and 2. This configuration works because any arc containing half the points will have at least half of them colored with either color 1 or color 2.

2. **Proof that \( k \leq 2 \):**
   - Assume for contradiction that \( k \geq 3 \).
   - By the Pigeonhole Principle, there must be a color that appears at most \( \left\lceil \frac{2021}{3} \right\rceil = 674 \) times. Let this color be color 1.
   - The number of points not colored by 1 is at least \( 2021 - 674 = 1347 \).

3. **Definition of Arc \( Y_i \):**
   - For each point \( a_i \) not colored by 1, define \( Y_i \) as the shortest arc containing \( a_i \) such that at least half of the points on it are colored by 1.

4. **Claim 1: One endpoint of \( Y_i \) is \( a_i \):**
   - Suppose the contrary, and let the endpoints of \( Y_i \) be \( m \) and \( n \).
   - Let \( z_1 \) be the number of points between \( m \) and \( a_i \) (excluding \( a_i \)), and \( z_2 \) be the number of these points colored by 1.
   - Similarly, let \( t_1 \) be the number of points between \( n \) and \( a_i \) (excluding \( a_i \)), and \( t_2 \) be the number of these points colored by 1.
   - We know \( z_2 < \frac{z_1 + 1}{2} \) (otherwise, the arc from \( m \) to \( a_i \) would be shorter and suitable for \( a_i \)).
   - Thus, \( z_2 \leq \frac{z_1}{2} \).
   - Similarly, \( t_2 \leq \frac{t_1}{2} \).
   - Therefore, \( z_2 + t_2 \leq \frac{z_1 + t_1}{2} \).
   - However, we also know \( z_2 + t_2 \geq \frac{z_1 + t_1 + 1}{2} \), leading to a contradiction.

5. **Claim 2: The second endpoint of \( Y_i \) is colored by 1:**
   - Suppose the contrary, and let the second endpoint of \( Y_i \) be \( m \), which is not colored by 1.
   - Let \( z_1 \) be the number of points between \( m \) and \( a_i \) (both included), and \( z_2 \) be the number of these points colored by 1.
   - We know \( z_2 \geq \frac{z_1}{2} \).
   - Removing \( m \) from the arc \( Y_i \) results in a shorter arc, and since \( z_2 > \frac{z_1 - 1}{2} \), the new arc still satisfies the condition, leading to a contradiction.

6. **Claim 3: Exactly half of the points on \( Y_i \) are colored by 1:**
   - Suppose the contrary, and let \( z_1 \) be the number of points between \( a_i \) and \( b_i \) (both included), and \( z_2 \) be the number of these points colored by 1.
   - If \( z_2 > \frac{z_1}{2} \), then \( z_2 \geq \frac{z_1 + 1}{2} \).
   - Removing \( b_i \) from the arc \( Y_i \) results in a shorter arc, and since \( z_2 - 1 \geq \frac{z_1 - 1}{2} \), the new arc still satisfies the condition, leading to a contradiction.

7. **Application of the Pigeonhole Principle:**
   - For each arc \( Y_i \), consider the point \( b_i \). All \( b_i \) points are colored by 1 (Claim 2).
   - There are at most 674 points colored by 1 and at least 1347 arcs.
   - By the Pigeonhole Principle, there must be at least 3 arcs with the same endpoint.
   - WLOG, let \( Y_m, Y_n, Y_p \) have the same endpoint \( b_1 \).
   - These arcs are on the same circle and end at the same point, so there must be 2 arcs where one covers the other.
   - WLOG, let \( Y_m \) cover \( Y_n \). Let \( a_m \) and \( a_n \) be their starting points.
   - Let \( z_1 \) be the number of points between \( a_m \) and \( a_n \) (excluding \( a_n \)), and \( z_2 \) be the number of these points colored by 1.
   - Let \( t_1 \) be the number of points between \( a_n \) and \( b_1 \) (both included), and \( t_2 \) be the number of these points colored by 1.
   - By Claim 3, \( t_2 = \frac{t_1}{2} \) (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 starting at \( a_m \) and ending just before \( a_n \) (arc \( Y_m - Y_n \)).
   - On this arc, there are \( z_1 \) points and \( z_2 \) of them are colored by 1.
   - Since \( z_2 = \frac{z_1}{2} \), this arc is suitable for \( a_m \), contradicting the assumption that \( Y_i \) was the shortest arc suitable for \( a_m \).

8. **Conclusion:**
   - Therefore, \( k \leq 2 \).

The final answer is \( \boxed{ k = 2 } \).