5. 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 form the vertices of an isosceles trapezoid (a convex quadrilateral with one pair of parallel sides and the other two sides equal in length but not parallel is called an isosceles trapezoid).
(Cold Gangsong, Contributed)
5. The smallest value of $n$ sought is 17.

First, prove: When $n=17$, the conclusion holds.
Proof by contradiction.
Assume 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 exist a set of 6 vertices of the same color. Without loss of generality, assume these are yellow. Connecting these 6 points pairwise, 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 scenarios must occur:
(1) There are three line segments of the same length.
Note that $3 \times 17$ does not allow these three line segments to share a common vertex. Therefore, there exist two line segments with no common vertices. The four 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 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 requirements. According to the pigeonhole principle, there must be two pairs of line segments that share the same yellow vertex. The other four vertices of these four line segments must form the vertices of an isosceles trapezoid, leading to a contradiction.
Therefore, when $n=17$, the conclusion holds.
Next, construct a coloring method for $n \leqslant 16$ that does not satisfy the problem's requirements. Let $A_{1}, A_{2}, \cdots, A_{n}$ represent the vertices of the regular $n$-gon (in clockwise order), and $M_{1}, M_{2}, M_{3}$ represent the sets of vertices of the three colors.
$$
\begin{array}{l}
\text { When } n=16 \text {, let } \\
M_{1}=\left\{A_{5}, A_{8}, A_{13}, A_{14}, A_{16}\right\}, \\
M_{2}=\left\{A_{3}, A_{6}, A_{7}, A_{11}, A_{15}\right\}, \\
M_{3}=\left\{A_{1}, A_{2}, A_{4}, A_{9}, A_{10}, A_{12}\right\} .
\end{array}
$$

For $M_{1}$, the distances from $A_{14}$ to the other 4 vertices are all different, and these 4 vertices form the vertices of a rectangle, not an isosceles trapezoid. For $M_{3}$, the 6 vertices are the endpoints of 3 diameters, so any 4 vertices either form the vertices of a rectangle or a non-isosceles quadrilateral.
$$
\begin{array}{l}
\text { When } n=15 \text {, let } \\
M_{1}=\left\{A_{1}, A_{2}, A_{3}, A_{5}, A_{8}\right\}, \\
M_{2}=\left\{A_{6}, A_{9}, A_{13}, A_{14}, A_{15}\right\}, \\
M_{3}=\left\{A_{4}, A_{7}, A_{10}, A_{11}, A_{12}\right\},
\end{array}
$$

Each $M_{i}$ contains no 4 points that form the vertices of an isosceles trapezoid.
$$
\begin{array}{l}
\text { When } n=14 \text {, let } \\
M_{1}=\left\{A_{1}, A_{3}, A_{8}, A_{10}, A_{14}\right\}, \\
M_{2}=\left\{A_{4}, A_{5}, A_{7}, A_{11}, A_{12}\right\}, \\
M_{3}=\left\{A_{2}, A_{6}, A_{9}, A_{13}\right\},
\end{array}
$$

Each $M_{i}$ contains no 4 points that form the vertices of an isosceles trapezoid.
When $n=13$, let
$$
\begin{array}{l}
M_{1}=\left\{A_{5}, A_{6}, A_{7}, A_{10}\right\}, \\
M_{2}=\left\{A_{1}, A_{8}, A_{11}, A_{12}\right\}, \\
M_{3}=\left\{A_{2}, A_{3}, A_{4}, A_{9}, A_{13}\right\},
\end{array}
$$

Each $M_{i}$ contains no 4 points that 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 removing vertex $A_{11}$ gives a coloring method for $n=10$.

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

The examples constructed above show that $n \leqslant 16$ does not satisfy the problem's requirements.
In conclusion, the smallest value of $n$ sought is 17.