25. Find the smallest positive integer $n$ such that: If each vertex of a regular $n$-gon is arbitrarily colored with one of the three colors red, yellow, or blue, then among the $n$ vertices, there must exist four vertices of the same color that form the vertices of an isosceles trapezoid (a convex quadrilateral with two parallel sides and two non-parallel sides of equal length is called an isosceles trapezoid).
25. The minimum value of $n$ sought is 17. First, we prove: when $n=17$, the conclusion holds. Suppose, for contradiction, there exists a method to three-color the vertices of a regular 17-gon such that no four vertices of the same color form the vertices of an isosceles trapezoid. Since $\left[\frac{17-1}{3}\right]+1=6$, there must be 6 vertices of the same color, say yellow. Connecting these 6 vertices pairwise, we get $C_{6}^{2}=15$ line segments. Since these line segments can have only $\left[\frac{17}{2}\right]=8$ different lengths, one of the following two scenarios must occur: (1) There are three line segments of the same length. Note that $3 \times 17$. It is impossible for these three line segments to have pairwise common points. Therefore, there exist two line segments with distinct endpoints. These four endpoints form the vertices of an isosceles trapezoid, a contradiction; (2) There are 7 pairs of line segments of the same length. By assumption, each pair of line segments of the same length must share a common yellow vertex (otherwise, we could find four yellow vertices that satisfy the problem's requirement). Since there are only 6 yellow vertices, by the pigeonhole principle, among these 7 pairs of line segments, there must be $\left[\frac{7-1}{6}\right]+1=2$ pairs of line segments sharing the same yellow vertex. The other four endpoints of these four line segments must form the vertices of an isosceles trapezoid, a contradiction. Therefore, the conclusion holds when $n=17$. Next, for $n \leqslant 16$, construct a coloring method that does not satisfy the problem's requirement. Let $A_{1}, A_{2}, \cdots, A_{n}$ represent the vertices of the regular $n$-gon (in clockwise order). $M_{1}, M_{2}, M_{3}$ represent the sets of vertices of the three colors. When $n=16$, $M_{1}=\left\{A_{5}, A_{8}, A_{13}, A_{14}, A_{16}\right\}, M_{2}=\left\{A_{3}, A_{6}, A_{7}, A_{11}, A_{15}\right\}, M_{3}=\left\{A_{1}, A_{2}, A_{4}, A_{9}, A_{10}\right.$, $\left.A_{12}\right\}$; when $n=15$, $M_{1}=\left\{A_{1}, A_{2}, A_{3}, A_{5}, A_{8}\right\}, M_{2}=\left\{A_{6}, A_{9}, A_{13}, A_{14}, A_{15}\right\}, M_{3}=\left\{A_{4}, A_{7}, A_{10}, A_{11}\right.$, $\left.A_{12}\right\}$; when $n=14$, let $M_{1}=\left\{A_{1}, A_{3}, A_{8}, A_{10}, A_{14}\right\}, M_{2}=\left\{A_{4}, A_{5}, A_{7}, A_{11}, A_{12}\right\}, M_{3}=\left\{A_{2}, A_{6}, A_{9}, A_{13}\right\}$; when $n=13$, $M_{1}=\left\{A_{5}, A_{6}, A_{7}, A_{10}\right\}, M_{2}=\left\{A_{1}, A_{8}, A_{11}, A_{12}\right\}, M_{3}=\left\{A_{2}, A_{3}, A_{4}, A_{9}, A_{13}\right\}$. For the case $n=13$, removing vertex $A_{13}$ and keeping the rest of the coloring unchanged, we get the coloring method for $n=12$; then removing vertex $A_{12}$, we get the coloring method for $n=11$; continuing to remove vertex $A_{11}$, we get the coloring method for $n=10$, and it is not difficult to verify that the above coloring methods ensure that no four vertices of the same color form the vertices of an isosceles trapezoid. When $n \leqslant 9$, we can ensure that the number of vertices of each color is less than 4, thus no four vertices of the same color form the vertices of an isosceles trapezoid. Therefore, when $n \leqslant 16$, the required property is not satisfied. In summary, the minimum value of $n$ sought is 17.