Initially $20^{\circ}$. There are 800 points on the circumference, labeled in a clockwise direction as $1,2, \cdots, 800$. They divide the circumference into 800 gaps. Now, choose one point and color it red, then follow the rule to color other points red one by one: if the $k$-th point has been colored red, then move $k$ gaps in a clockwise direction and color the endpoint reached red. Continue this process. How many red points can be obtained at most on the circumference? Prove your conclusion.
Proof Consider a circle with $2n$ points in general.
(1) On a circle with $2n$ points, if the first red point is an even-numbered point, for example, the $2k$-th point, then according to the coloring rule, each red point dyed afterward will also be an even-numbered point. At this time, if the points numbered $2, 4, 6, \cdots, 2k, \cdots, 2n$ are renumbered as $1^{*}, 2^{*}, 3^{*}, \cdots, k^{*}, \cdots, n^{*}$, it is easy to see that the number of red points that can be dyed starting from the $2k$-th point on a circle with $2n$ points is the same as the number of red points that can be dyed starting from the $k$-th point on a circle with $n$ points, and they form a one-to-one correspondence.
(2) On a circle with $2n$ points, if the first red point is an odd-numbered point, then the second red point must be an even-numbered point. By (1), in this case, it can also correspond to a coloring method starting from the $k$-th point on a circle with $n$ points, and the number of red points dyed is exactly one more than the number of red points obtained by this coloring method on the circle with $n$ points.
(3) Let the maximum number of red points that can be dyed on a circle with $n$ points be $f(n)$, and suppose the coloring method starting from the $k$-th point achieves this maximum value. Then, on a circle with $2n$ points, the coloring method starting from the $k$-th point should achieve the maximum value $f(n)+1$ (since the second red point is the $2k$-th point). Therefore, $f(2n)=f(n)+1$.
(4) From (3), we have
$$
\begin{array}{l}
f(800)=f(400)+1=f(200)+2 \\
=f(100)+3=f(50)+4=f(25)+5 .
\end{array}
$$
(5) To calculate $f(25)$, note that on a circle with 25 points, if the number of the first red point is a multiple of 5, then the number of each red point dyed afterward will also be a multiple of 5. Therefore, the number of red points that can be dyed starting from such a point will not exceed 5.

If the number of the first red point is not a multiple of 5, then the number of each red point dyed afterward will also not be a multiple of 5. Therefore, starting from such a point, at most 20 red points can be dyed.

On the other hand, starting from a certain such point, we can actually dye 20 red points, for example, starting from the 1st point, we can sequentially dye 1, 2, 4, 8, 16, 7, 14, 3, 6, 12, 24, 23, 21, 17, 9, 18, 11, 22, 19, 13. Therefore, $f(25)=20$.

From (4) and (5), we get $f(800)=20+5=25$, i.e., the maximum number of red points that can be dyed is 25.
(Jiangxi Nanchang Vocational and Technical Normal University, 330013, Gou Ping Sheng's solution)