1. Let the side length of the equilateral $\triangle A B C$ be $k\left(k \in \mathbf{Z}_{+}\right.$, $k \geqslant 3)$, and divide $\triangle A B C$ into $k^{2}$ unit equilateral triangles (Figure 1 shows the case when $k=7$). Fill each unit equilateral triangle with one of the $k^{2}$ positive integers from $1$ to $k^{2}$. Six unit equilateral triangles that share a common vertex form a unit hexagon, and the sum of the numbers in these six unit equilateral triangles is recorded as the “area” of this unit hexagon. Find the maximum possible value of the sum of the areas of all unit hexagons (expressed in terms of $k$).
1. Classify the number of unit equilateral triangles that belong to a unit regular hexagon.
(1) Note that, the triangle with vertices $A, B, C$ does not belong to a unit regular hexagon. Fill $\{1,2,3\}$ into the unit equilateral triangle with vertices $A, B, C$, and denote its sum as $S_{1}$. Then
$$
S_{1}=6 \text {. }
$$
(2) Unit equilateral triangles that belong to only one unit regular hexagon.

Note that, in $\triangle A B C$, there are $k-2$ unit equilateral triangles with one of its sides as the base, totaling $3(k-2)=3 k-6$ triangles for the three sides. Unit equilateral triangles adjacent to the vertices $A, B, C$ also belong to only one unit regular hexagon. Thus, fill $\{4,5, \cdots, 3 k\}$ into such unit equilateral triangles, and denote their sum as $S_{2}$. Then
$$
s_{2}=4+5+\cdots+3 k=\frac{(3 k-3)(3 k+4)}{2} \text {. }
$$
(3) Unit equilateral triangles that belong to exactly two unit regular hexagons.

Note that, in $\triangle A B C$, there are $k-3$ unit equilateral triangles with one of its sides' points as the vertex, totaling $3(k-3)=3 k-9$ triangles for the three sides. Fill $\{3 k+1,3 k+2, \cdots, 6 k-9\}$ into such unit equilateral triangles, and denote their sum as $S_{3}$, which is
$$
\begin{array}{l}
s_{3}=(3 k+1)+(3 k+2)+\cdots+(6 k-9) \\
=\frac{(3 k+1)+(6 k-9)}{2}(3 k-9) .
\end{array}
$$
(4) Unit equilateral triangles that belong to three unit regular hexagons.

These triangles are the unit equilateral triangles in $\triangle D E F$, as shown in Figure 2, totaling $(k-3)^{2}=k^{2}-6 k+9$ triangles.
Fill $\left\{6 k-8,6 k-7, \cdots, k^{2}\right\}$ into such unit equilateral triangles, and denote their sum as $S_{4}$, which is
$$
\begin{array}{l}
s_{4}=(6 k-8)+(6 k-7)+\cdots+k^{2} \\
=\frac{k^{2}+6 k-8}{2}\left(k^{2}-6 k+9\right) . \\
\text { Hence } s=s_{2}+2 s_{3}+3 s_{4} \\
=\frac{3\left(k^{4}-14 k^{2}+33 k-28\right)}{2} .
\end{array}
$$

Thus, $s=s_{2}+2 s_{3}+3 s_{4}$

Next, we prove that $S$ in equation (1) is the maximum value.
Let $x \in S_{3}, y \in S_{4}$.
Then $x3 x+2 y \text {. }$
$$

Therefore, the maximum value is
$$
\frac{3\left(k^{4}-14 k^{2}+33 k-28\right)}{2} \text {. }
$$