Problem: Let $(m,n)$ be positive integers with $n \ge 3$ 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?

Solution: The answer is if and only if $m$ is a proper divisor of $n$.

Throughout this solution, we let $\omega = \exp\left( 2 \pi i / n \right)$
and let the regular $n$-gon have vertices $1$, $\omega$, \dots, $\omega^{n-1}$.
We cache the following frequent calculation:
\begin{lemma*}
  The triangle with vertices $\omega^k$, $\omega^{k+a}$, $\omega^{k+b}$ has signed area
  \[ T(a,b) \coloneq \frac{(\omega^a-1)(\omega^b-1)(\omega^{-b}-\omega^{-a})}{4i}. \]
\end{lemma*}
\begin{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
  \[
    \frac{i}{4}
    \det \begin{bmatrix}
      1 & 1 & 1 \\
      \omega^a & \omega^{-a} & 1 \\
      \omega^b & \omega^{-b} & 1 \\
    \end{bmatrix}
    =
    \frac{i}{4}
    \det \begin{bmatrix}
      0 & 0 & 1 \\
      \omega^a-1 & \omega^{-a}-1 & 1 \\
      \omega^b-1 & \omega^{-b}-1 & 1 \\
    \end{bmatrix}
  \]
  which equals the above.
\end{proof}

\paragraph{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:
\begin{center}
\begin{asy}
  size(6cm);
  pair A(int i) { return dir(40*i); }
  pen[] fillcolors = {palered, palegreen, palecyan};
  pen[] labelcolors = {brown, darkgreen, deepblue};
  string[] labels = {"R", "Green", "Blue", "Red", "Green", "Blue", "R"};
  for (int i=0; i<=6; ++i) {
    filldraw(A(0)--A(i+1)--A(i+2)--cycle, fillcolors[i-3*(i#3)], black+1.5);
    label(labels[i], incenter(A(0),A(i+1),A(i+2)),
      labelcolors[i-3*(i#3)] + fontsize(14pt));
  }
\end{asy}
\end{center}
To see this works one can just do the shoelace calculation:
fix a residue $r \bmod m$ corresponding to one of the colors.
Then
\begin{align*}
  \sum_{\substack{0 \le j < n \\ j \equiv r \bmod m}} \operatorname{Area}(\omega^0, \omega^j, \omega^{j+1})
  &= \sum_{\substack{0 \le j < n \\ j \equiv r \bmod m}} T(j, j+1) \\
  &= \sum_{\substack{0 \le j < n \\ j \equiv r \bmod m}}
    \frac{(\omega^j-1)(\omega^{j+1}-1)(\omega^{-(j+1)}-\omega^{-j})}{4i} \\
  &= \frac{\omega-1}{4i} \sum_{\substack{0 \le j < n \\ j \equiv r \bmod m}}
    (\omega^{-j}-1)(\omega^j-\omega^{-1}) \\
  &= \frac{\omega-1}{4i} \left( \frac{n}{m} \left( 1 + \omega\inv \right)
    + \sum_{\substack{0 \le j < n \\ j \equiv r \bmod m}} (\omega^{-1-j}-\omega^j) \right).
\end{align*}
(We allow degenerate triangles where $j \in \{0,m-1\}$ with area zero
to simplify the notation above.)

However, if $m$ is a proper divisor of $m$, then
$\sum_j \omega^j = \omega^r(1+\omega^m+\omega^{2m}+\dots+\omega^{n-m}) = 0$.
For the same reason,
$\sum_j \omega^{-1-j} = 0$.
So the inner sum vanishes, and the total area of this color equals
\[ \sum_{\substack{0 \le j < n \\ j \equiv r \bmod m}}
  \operatorname{Area}(\omega^0, \omega^j, \omega^{j+1})
  = \frac nm \frac{(\omega-1)(\omega^{-1}+1)}{4i}
  = \frac nm \cdot \frac{\omega-\omega^{-1}}{4i}
. \]
Because the right-hand side does not depend on the residue $r$,
this shows all colors have equal area.

\paragraph{Proof of necessity.}
It's obvious that $m < n$ (in fact $m \le n-2$).
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
\[ S \coloneq \frac nm \cdot \frac{\omega-\omega^{-1}}{4i}. \]
However:
\begin{claim*}
  The number $4i \cdot S$ is not an algebraic integer when $m \nmid n$.
\end{claim*}
\begin{proof}
  This is easiest to see if one knows the advanced result that
  $K \coloneq \QQ(\omega)$ is a number field
  whose ring of integers is known to be $\mathcal{O}_K = \ZZ[\omega]$,
  in which case it follows right away.
\end{proof}
\begin{remark*}
  We spell out the details in the proof a bit more explicitly here.
  It's enough to show that
  $\omega \cdot 4i \cdot S = \frac nm \omega^2 - \frac nm$
  is not an algebraic integer for completeness.

  Take $K = \mathbb Q(\omega)$ of degree $d \coloneq \varphi(n) \ge 2$;
  as a $\mathbb Q$-module, it obeys
  $K = \mathbb Q \oplus \omega \mathbb Q \oplus \dots \oplus \omega^{d-1} \mathbb Q$.
  The theorem we are quoting is that, as $\ZZ$-modules, we have
  $\mathcal{O}_K = \mathbb Z[\omega] =
    \mathbb Z \oplus \omega \mathbb Z \oplus \dots \oplus \omega^{d-1} \mathbb Z$
  i.e.\ $\mathcal{O}_K$ contains exactly those numbers in $K$
  for which the canonical $\mathbb Q$-coefficients happen to be integers.
  (This is quite famous, but for a print reference,
  see Neukirch's \emph{Algebraic Number Theory}, Section I.10.)
  And $\omega \cdot 4i \cdot S$ fails this criteria, since $\frac nm \notin \ZZ$.
\end{remark*}
However, for any $a$ and $b$, the number
\[ 4i \cdot T(a,b) = (\omega^a-1)(\omega^b-1)(\omega^{-b}-\omega^{-a}) \]
is an algebraic integer.
Since a finite sum of algebraic integers is also an algebraic integer,
the sum of expressions of the form $4i \cdot T(a,b)$ will never equal $4i \cdot S$.

\begin{remark*}
  If one wants to avoid citing the fact that $\mathcal{O}_K = \ZZ[\omega]$,
  then one can instead note that $T(a,b)$ is actually always divisible by
  $(\omega-1)(\omega^{-1}+1) = \omega - \omega^{-1}$ over the algebraic integers
  (at least one of $\{\omega^a-1, \omega^b-1, \omega^{-a} - \omega^{-b}\}$
  is a multiple of $\omega+1$, by casework on $a,b \bmod 2$).
  Then one using $\frac{4i}{(\omega-1)(\omega^{-1}+1)}$ as the scaling factor
  instead of $4i$, one sees that we actually need $\frac nm$ to be an algebraic integer.
  But a rational number is an algebraic integer if and only if it's a rational integer;
  so $\frac nm$ is an algebraic integer exactly when $m$ divides $n$.
\end{remark*}