100 points on a circle divide it into 100 equal arcs. Randomly color some of these points red, ensuring that at least 4 red points are the vertices of a square. How many points do you need to color red at minimum?

To ensure that at least 4 red points form the vertices of a square, you need to consider the distribution of the points. The circle is divided into 100 equal segments, and a square can be formed by selecting 4 points that are equally spaced around the circle. 

To find the minimum number of red points required, we can use the pigeonhole principle. If we want to avoid having 4 red points forming a square, we can distribute the red points in such a way that no 4 points form a square. 

However, if we color more points, the likelihood of forming a square increases. The key is to find the minimum number of points that guarantees a square will be formed.

The solution involves considering the worst-case scenario where we try to avoid forming a square. If we color 25 points, we can place them such that no 4 points form a square. However, if we color 26 points, it becomes impossible to avoid forming a square.

Therefore, the minimum number of points that need to be colored red to ensure that at least 4 red points form the vertices of a square is 26.
25. Answer: 76

Analysis: As shown in the figure: when a pair of diameters $\mathrm{AC}, \mathrm{BD}$ of a circle are perpendicular to each other, then $\mathrm{ABCD}$ is precisely a square. Conversely, if four points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ on the circle are exactly the four vertices of a square $\mathrm{ABCD}$, then the diagonals $\mathrm{AC}, \mathrm{BD}$ are precisely a pair of mutually perpendicular diameters of the circle.

100 points on the circle divide it into 100 equal arcs, forming exactly 25 pairs of mutually perpendicular diameters. The 4 endpoints of each pair of mutually perpendicular diameters can form 25 different squares. In the least favorable scenario: 3 points out of the 4 endpoints of each pair of mutually perpendicular diameters are colored red, resulting in a total of 75 points being colored red among the 100 equal division points on the circle, where no 4 red points form the vertices of a square. At this point, by coloring one more red point, i.e., coloring 76 points red, and since $76 = 3 \times 25 + 1$, there will definitely be a square whose 4 vertices are all red points. Therefore, to ensure that at least one square has all 4 vertices colored red, at least 76 of the 100 equal division points must be colored red.