Let $k$ be a positive integer. Each cell of a $3 k \times 3 k$ grid is colored either red or blue. A triple of cells $\left(C_{1}, C_{2}, C_{3}\right)$ is called a good cell group if $C_{1}$ and $C_{2}$ are in the same row, $C_{2}$ and $C_{3}$ are in the same column, and $C_{1}, C_{3}$ are colored blue, while $C_{2}$ is colored red. Find the maximum possible number of good cell groups.
Let the $i$-th row have $a_{i}$ blue small squares, and the $j$-th column have $b_{j}$ blue small squares $(1 \leqslant i \leqslant 3 k, 1 \leqslant j \leqslant 3 k)$. Let $R$ be the set of all red small squares, and use $(i, j)$ to denote the small square in the $i$-th row and $j$-th column. For each red small square $C_{2}(i, j)$, there are $a_{i} b_{j}$ good square groups $\left(C_{1}, C_{2}, C_{3}\right)$ that satisfy the conditions. Let the number of good square groups be $T$, then
$$
T=\sum_{(i, j) \in R} a_{i} b_{j} \leqslant \frac{1}{2} \sum_{(i, j) \in R}\left(a_{i}^{2}+b_{j}^{2}\right)=\frac{1}{2} \sum_{i=1}^{3 k}\left(3 k-a_{i}\right) a_{i}^{2}+\frac{1}{2} \sum_{j=1}^{3 k}\left(3 k-b_{j}\right) b_{j}^{2} .
$$

This is because the $i$-th row has $3 k-a_{i}$ red small squares, and the $j$-th column has $3 k-b_{j}$ red small squares.
For $x \in\{0,1,2, \cdots, 3 k\}$, by the mean inequality, we get
$$
(3 k-x) x^{2}=\frac{1}{2}(6 k-2 x) \cdot x \cdot x \leqslant \frac{1}{2}\left(\frac{1}{3}((6 k-2 x)+x+x)\right)^{3}=4 k^{3} .
$$

In the above formula, let $x=a_{i}, x=b_{j}$, substitute into formula (1), we get
$$
T \leqslant 12 k^{4} .
$$

When the equality in the above formula holds, the inequalities (1), (2) should both take equality, then
$$
\begin{array}{l}
6 k-2 a_{i}=a_{i} \Rightarrow a_{i}=2 k, \\
6 k-2 b_{j}=b_{j} \Rightarrow b_{j}=2 k,
\end{array}
$$

where $i, j \in\{1,2, \cdots, 3 k\}$.
Under $\bmod 3 k$, when $i-j \in\{1,2, \cdots, 2 k\}$, color the square $(i, j)$ blue, and color the other squares red, i.e., when $i$ is fixed, each row has $2 k$ blue squares, and when $j$ is fixed, each column also has $2 k$ blue squares. Thus, $T$ can reach the maximum value $12 k^{4}$.