Example 5 For which $n \in \mathbf{N}^{\cdot}, n \geqslant 5$, can the vertices of a regular $n$-gon be colored using no more than 6 colors, such that any 5 consecutive vertices have distinct colors?
Let the colors be $a, b, c, d, e, f$. Define the sequence $A: a, b, c, d, e$ and the sequence $B: a, b, c, d, e, f$.

If there exist non-negative integers $x, y$, such that $n=5 x+6 y$, then for a regular $n$-gon, the vertices can be colored by first coloring $y$ sequences of $B$, followed by $x$ sequences of $A$, ensuring that any 5 consecutive vertices are of different colors.

Using the conclusion from Example 3, we know that when $n \geqslant 5 \times 6-(5+6)+1=20$, the equation $n=5 x+6 y$ always has non-negative integer solutions. For $5 \leqslant n \leqslant 19$, direct calculation shows that the equation $n=5 x+6 y$ has no non-negative integer solutions only when $n \in \{7,8,9,13,14,19\}$.

On the other hand, for $n \in \{7,8,9,13,14,19\}$, there exists $k \in \mathbf{N}^{*}$ such that $6 k<n<6(k+1)$. Therefore, there must be a color that appears $k+1$ times. Since these $k+1$ points of the same color are at least 4 points apart, we have
$$n \geqslant 5(k+1)$$

Now,
when $n \in \{7,8,9\}$, $k=1$, requiring $n \geqslant 10$, which is a contradiction.
when $n \in \{13,14\}$, $k=2$, requiring $n \geqslant 15$, which is a contradiction.
when $n=19$, $k=3$, requiring $n \geqslant 20$, which is also a contradiction.
In summary, when $n \geqslant 5$, except for the numbers in the set $\{7,8,9,13,14,19\}$, all other positive integers meet the requirement.