To solve this problem, we need to understand the conditions under which an $m$-balanced triangulation of a regular $n$-gon is possible. The concept of $m$-balanced means that each color covers exactly the same total area across all triangles of that color. Here's a breakdown of the solution:

Consider a regular $n$-gon, and let's triangulate it. The total area of the $n$-gon is equally divided among the triangles formed. When coloring these triangles with $m$ colors in an $m$-balanced manner, each color must cover $\frac{1}{m}$ of the total polygon area.

Key Conditions:

1. Each triangle in the triangulation has an equal area because the $n$-gon is regular.
2. The total number of triangles that can be formed in a regular $n$-gon triangulation is $n - 2$.
3. For the triangulation to be $m$-balanced, the total number of triangles $n - 2$ has to be divisible by $m$, i.e., $m \mid (n-2)$.

Additionally, because a regular polygon of $n$ sides can only be triangulated if $n \geq 3$, we have $n \geq 3$. Furthermore, to be able to partition these triangles into $m$ groups of equal area, clearly $n$ needs to be larger than $m$, leading to the second condition $n > m$.

Final Condition:

Combining these conditions, we conclude:

- $m \mid (n-2)$
- $n > m$
- $n \geq 3$

Thus, the set of all positive integers $n$ for which there exists an $m$-balanced triangulation of a regular $n$-gon is characterized by:

$$\boxed{m \mid n \text{ with } n > m \text{ and } n \geq 3}$$

This conclusion provides a comprehensive characterization of all such $n$ where an $m$-balanced triangulation exists for a given $m$-gon.