Let $(m, n)$ be positive integers with $n \geq 3$ and draw a regular $n$-gon. We wish to triangulate this $n$-gon into $n-2$ triangles, each colored one of $m$ colors, so that each color has an equal sum of areas. For which $(m, n)$ is such a triangulation and coloring possible?
The answer is if and only if $m$ is a proper divisor of $n$.
Throughout this solution, we let $\omega=\exp (2 \pi i / n)$ and let the regular $n$-gon have vertices $1, \omega, \ldots, \omega^{n-1}$. We cache the following frequent calculation:
Lemma
The triangle with vertices $\omega^{k}, \omega^{k+a}, \omega^{k+b}$ has signed area
$$
T(a, b):=\frac{\left(\omega^{a}-1\right)\left(\omega^{b}-1\right)\left(\omega^{-a}-\omega^{-b}\right)}{2 i} .
$$

Proof. Rotate by $\omega^{-k}$ to assume WLOG that $k=0$. Apply complex shoelace to the triangles with vertices $1, \omega^{a}, \omega^{b}$ to get
$$
\frac{1}{2 i} \operatorname{det}\left[\begin{array}{ccc}
1 & 1 & 1 \\
\omega^{a} & \omega^{-a} & 1 \\
\omega^{b} & \omega^{-b} & 1
\end{array}\right]=\frac{1}{2 i} \operatorname{det}\left[\begin{array}{ccc}
0 & 0 & 1 \\
\omega^{a}-1 & \omega^{-a}-1 & 1 \\
\omega^{b}-1 & \omega^{-b}-1 & 1
\end{array}\right]
$$
which equals the above.
『 Construction. It suffices to actually just take all the diagonals from the vertex 1 , and then color the triangles with the $m$ colors in cyclic order. For example, when $n=9$ and $m=3$, a coloring with red, green, blue would be:
To see this works one can just do the shoelace calculation: for a given residue $r \bmod m$, we get an area
$$
\begin{aligned}
\sum_{j \equiv r \bmod m} \operatorname{Area}\left(\omega^{j}, \omega^{0}, \omega^{j+1}\right) & =\sum_{j \equiv r \bmod m} T(-j, 1) \\
& =\sum_{j \equiv r \bmod m} \frac{\left(\omega^{-j}-1\right)\left(\omega^{1}-1\right)\left(\omega^{j}-\omega^{-1}\right)}{2 i} \\
& =\frac{\omega-1}{2 i} \sum_{j \equiv r \bmod m}\left(\omega^{-j}-1\right)\left(\omega^{j}-\omega^{-1}\right) \\
& =\frac{\omega-1}{2 i}\left(\frac{n}{m}\left(1+\omega^{-1}\right)+\sum_{j \equiv r \bmod m}\left(\omega^{-j}-\omega^{j}\right)\right) .
\end{aligned}
$$
(We allow degenerate triangles where $j \in\{-1,0\}$ with area zero.) However, if $m$ is a proper divisor of $m$, then $\sum_{j \equiv r \bmod m} \omega^{j}=\omega^{r}\left(1+\omega^{m}+\omega^{2 m}+\cdots+\omega^{n-m}\right)=0$. Similarly, $\sum_{j \equiv r \bmod m} \omega^{-j}=0$. So the inner sum vanishes, and the total area of the $m$ th color equals
$$
\frac{n}{m} \frac{(\omega-1)\left(\omega^{-1}+1\right)}{2 i}
$$
which does not depend on the residue $r$, proving the coloring works.
『 Proof of necessity. It's obvious that $m<n$ (in fact $m \leq n-2$ ). So we focus on just showing $m \mid n$.

Repeating the same calculation as above, we find that if there was a valid triangulation and coloring, the total area of each color would equal
$$
S:=\frac{n}{m} \frac{(\omega-1)\left(\omega^{-1}+1\right)}{2 i} .
$$

However:
Claim - The number $2 i \cdot S$ is not an algebraic integer when $m \nmid n$.

Proof. This is easiest to see if one knows the advanced result that $K:=\mathbb{Q}(\omega)$ is a number field whose ring of integers is known to be $\mathcal{O}_{K}=\mathbb{Z}[\omega]$. Hence if one takes $\left(\omega^{-1}, \omega^{0}, \omega^{1}, \ldots, \omega^{n-2}\right)$ as a $\mathbb{Q}$-basis of $K$, then $\mathcal{O}_{K}$ is the subset where each coefficient is integer.

However, each of the quantities $T(a, b)$ is $\frac{1}{2 i}$ times an algebraic integer. Since a finite sum of algebraic integers is also an algebraic integer, such areas can never sum to $S$.
Remark. If one wants to avoid citing the fact that $\mathcal{O}_{K}=\mathbb{Z}[\omega]$, then one can instead note that $T(a, b)$ is actually always divisible by $(\omega-1)\left(\omega^{-1}+1\right)$ over the algebraic integers (at least one of $\left\{\omega^{a}-1, \omega^{b}-1, \omega^{-a}-\omega^{-b}\right\}$ is a multiple of $\omega+1$, by casework on $a, b \bmod 2$ ). Then one using $\frac{2 i}{(\omega-1)\left(\omega^{-1}+1\right)}$ as the scaling factor instead of $2 i$, one sees that we actually need $\frac{n}{m}$ to be an algebraic integer, which happens only when $m$ divides $n$.