6. 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 there must exist four vertices of the same color that are the vertices of an isosceles trapezoid. (2008 China Mathematical Olympiad)
6. First construct a coloring method that does not meet the problem's requirements for $n \leqslant 16$.

Let $A_{1}, A_{2}, \cdots, A_{n}$ represent the vertices of a regular $n$-sided polygon (in clockwise order), and $M_{1}, M_{2}, M_{3}$ represent the sets of vertices of three different colors.

When $n=16$, let $M_{1}=\left\{A_{5}, A_{8}, A_{13}, A_{14}, A_{16}\right\}, M_{2}=\left\{A_{3}, A_{6}, A_{7}, A_{11}, A_{15}\right\}, M_{3}=\left\{A_{1}, A_{2}, A_{4}, A_{9}, A_{10}, A_{12}\right\}$. For $M_{1}$, the distance from $A_{14}$ to the other 4 vertices is different, and these 4 vertices form a rectangle. Similarly to $M_{1}$, it can be verified that $M_{2}$ does not contain 4 vertices that form the vertices of an isosceles trapezoid. For $M_{3}$, the 6 vertices are the endpoints of 3 diameters, so any 4 vertices either form the 4 vertices of a rectangle or the 4 vertices of a non-equilateral quadrilateral.

When $n=15$, let $M_{1}=\left\{A_{1}, A_{2}, A_{3}, A_{5}, A_{8}\right\}, M_{2}=\left\{A_{6}, A_{9}, A_{13}, A_{14}, A_{15}\right\}, M_{3}=\left\{A_{4}, A_{7}, A_{10}, A_{11}, A_{12}\right\}$, and no 4 points in each $M_{i}$ form the vertices of an isosceles trapezoid.

When $n=14$, let $M_{1}=\left\{A_{1}, A_{3}, A_{8}, A_{10}, A_{14}\right\}, M_{2}=\left\{A_{4}, A_{6}, A_{7}, A_{11}, A_{12}\right\}, M_{3}=\left\{A_{2}, A_{6}, A_{9}, A_{13}\right\}$, and no 4 points in each $M_{i}$ form the vertices of an isosceles trapezoid.

When $n=13$, let $M_{1}=\left\{A_{5}, A_{6}, A_{7}, A_{10}\right\}, M_{2}=\left\{A_{1}, A_{8}, A_{11}, A_{12}\right\}, M_{3}=\left\{A_{2}, A_{3}, A_{4}, A_{9}, A_{13}\right\}$, and no 4 points in each $M_{i}$ form the vertices of an isosceles trapezoid.

In the above cases, removing vertex $A_{13}$ and keeping the coloring method unchanged gives a coloring method for $n=12$; then removing vertex $A_{12}$ gives a coloring method for $n=11$; and continuing to remove vertex $A_{11}$ gives a coloring method for $n=10$.

When $n \leqslant 9$, the number of vertices of each color can be less than 4, so there are no 4 vertices of the same color that form the vertices of an isosceles trapezoid.

Therefore, $n \leqslant 16$ does not have the property required by the problem.
Next, we prove that $n=17$ satisfies the conclusion.
Proof by contradiction. Assume there exists a way to color the vertices of a regular 17-sided polygon with three colors such that no 4 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 points in pairs, we get $\mathrm{C}_{6}^{2}=15$ line segments. Since these line segments can have only $\left[\frac{17}{2}\right]=8$ different lengths, one of the following two cases must occur:
(1) There are 3 line segments of the same length.
Note that 3 < 17, so it is impossible for these 3 line segments to have a common vertex. Therefore, there must be two line segments with no common vertices. The 4 vertices of these two line segments satisfy the problem's requirements, leading to a contradiction.
(2) There are 7 pairs of line segments of the same length.

By the assumption, each pair of line segments of the same length must have a common yellow vertex, otherwise we could find 4 yellow vertices that satisfy the problem's requirements. By the pigeonhole principle, there must be two pairs of line segments with the same common yellow vertex. The other 4 vertices of these 4 line segments must form the vertices of an isosceles trapezoid, leading to a contradiction. Therefore, $n=17$ satisfies the conclusion.

In summary, the smallest value of $n$ is 17.