Example 2 Divide each side of the equilateral $\triangle A B C$ into $n$ equal parts, and draw lines parallel to the other two sides through the division points, resulting in a triangular grid array, or simply a triangular grid. How many rhombuses are there in the triangular grid?

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Let $n=2 m+\varepsilon(\varepsilon \in\{0,1\})$.
Using the method from Example 1, first consider the A-type rhombi formed by the intersection of $W_{b}$ and $W_{c}$.

As shown in Figure 1, take any A-type rhombus $D E I J$ with side length $k(k \leqslant m)$, where $D E, I J \in W_{b}$ and $E I, D J \in W_{c}$. Extend $D J, E I, J I, D E$ to intersect line $l$ at points $D^{\prime}, E^{\prime}, I^{\prime}, G^{\prime}$, respectively. Then $D^{\prime} E^{\prime}=I^{\prime} G^{\prime}=k$.

It is easy to see that there is a one-to-one correspondence between the rhombus $D E I J$ and the ordered pairs of points $\left(E^{\prime}, I^{\prime}\right)\left(E^{\prime} \neq A_{1}, A_{2}, \cdots, A_{k}, I^{\prime} \neq A_{n-k+3}, \cdots, A_{n+2}\right)$ on line $l$. Therefore, the number of A-type rhombi with side length $k(1 \leqslant k \leqslant m)$ is $g_{a}(n, k)=\mathrm{C}_{n+2-2 k}^{2}$.
Thus, the total number of A-type rhombi is
$$
\begin{array}{l}
g_{a}(n)=\sum_{k=1}^{m} \mathrm{C}_{n+2-2 k}^{2} \\
= \frac{1}{2} \sum_{k=1}^{m}(2 m+\varepsilon+2-2 k)(2 m+\varepsilon+1-2 k) \\
= 2 \sum_{k=1}^{m} k^{2}-(4 m+2 \varepsilon+3) \sum_{k=1}^{m} k+ \\
\frac{m}{2}(2 m+\varepsilon+2)(2 m+\varepsilon+1) \\
= \frac{m(m+1)(2 m+1)}{3}- \\
\frac{(4 m+2 \varepsilon+3) m(m+1)}{2}+ \\
\frac{m(2 m+\varepsilon+2)(2 m+\varepsilon+1)}{2} \\
= \frac{m}{6}\left[4 m^{2}+(6 \varepsilon+3) m+\left(3 \varepsilon^{2}+3 \varepsilon-1\right)\right] .
\end{array}
$$

Similarly, the number of B-type and C-type rhombi is the same.
Therefore, the total number of rhombi is
$$
g(n)=\frac{m}{2}\left[4 m^{2}+(6 \varepsilon+3) m+\left(3 \varepsilon^{2}+3 \varepsilon-1\right)\right],
$$

where $m=\left[\frac{n}{2}\right], \varepsilon=n-2 m$.