For which integers $n \ge 5$ is it possible to color the vertices of a regular$ n$-gon using at most $6$ colors in such a way that any $5$ consecutive vertices have different colors?
We need to determine for which integers \( n \ge 5 \) it is possible to color the vertices of a regular \( n \)-gon using at most 6 colors such that any 5 consecutive vertices have different colors.

1. **Case 1: \( n \) is of the form \( 5x \)**

   If \( n = 5x \), we can choose any five of the six colors and repeat them in the same order. For example, if we choose colors \( (a, b, c, d, e) \), we can color the vertices as \( (a, b, c, d, e, a, b, c, d, e, \ldots) \). This ensures that any 5 consecutive vertices have different colors.

2. **Case 2: \( n \) is of the form \( 5x + 6y \), where \( y > 0 \)**

   If \( n = 5x + 6y \), we can choose any five of the six colors, say \( (a, b, c, d, e) \), and repeat them in that same order \( x \) times. Then, we use \( y \) chains of all six colors in the same order \( (a, b, c, d, e, f) \). This ensures that any 5 consecutive vertices have different colors.

3. **Case 3: \( n = 7, 8, 9, 13, 14, 19 \)**

   These numbers are not represented in the forms \( 5x \) or \( 5x + 6y \). We will show that the coloring is not possible for these values of \( n \).

   Let \( n = 6k + l \), where \( l = 0, 1, 2, 3, 4, 5 \). Since \( n < 6(k + 1) \), by the Pigeonhole Principle, at least \( k + 1 \) vertices must be of the same color. This suggests that \( n > 5k + 5 \), which is not true for \( n = 7, 8, 9, 13, 14, 19 \). Therefore, it is not possible to color the vertices of a regular \( n \)-gon with these values of \( n \) using at most 6 colors such that any 5 consecutive vertices have different colors.

Thus, the final answer is the given range excluding only the set \(\{7, 8, 9, 13, 14, 19\}\).

The final answer is \( \boxed{ \{ n \ge 5 \mid n \notin \{7, 8, 9, 13, 14, 19\} \} } \)