1. A social club has $2 k+1$ members, each of whom is proficient in the same $k$ languages, and any two members can only communicate in one language. If there do not exist three members who can communicate pairwise in the same language, let $A$ be the number of subsets of three members, where each subset's three members communicate pairwise in different languages. Find the maximum value of $A$.

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1. The maximum value of $A$ is $\frac{2 k(k-2)(2 k+1)}{3}$.

Consider the social club as a complete graph with $2 k+1$ vertices, where each language corresponds to a color, and the edge connecting two members who speak the same language is colored with the corresponding color.

Let the set of vertices be $V$, and the set of languages or colors be $L=\left\{l_{1}, l_{2}, \cdots, l_{k}\right\}$. Let $d_{i}(v)$ denote the number of edges of color $l_{i}$ emanating from vertex $v$.
Then, for each vertex $v$, we have $\sum_{i=1}^{k} d_{i}(v)=2 k$.
If a triangle has two edges of the same color, it is called an "isosceles" triangle.

Next, by counting the number of pairs of edges of the same color sharing a vertex, the number of isosceles triangles sharing vertex $v$ is
$$
\mathrm{C}_{d_{1}(v)}^{2}+\mathrm{C}_{d_{2}(v)}^{2}+\cdots+\mathrm{C}_{d_{k}(v)}^{2} \text {. }
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\mathrm{C}_{d_{1}(v)}^{2}+\mathrm{C}_{d_{2}(v)}^{2}+\cdots+\mathrm{C}_{d_{k}(v)}^{2} \\
=\frac{1}{2} \sum_{i=1}^{k} d_{i}^{2}(v)-\frac{1}{2} \sum_{i=1}^{k} d_{i}(v) \\
\geqslant \frac{1}{2} \cdot \frac{\left(\sum_{i=1}^{k} d_{i}(v)\right)^{2}}{k}-\frac{1}{2} \times 2 k \\
=\frac{1}{2} \times \frac{(2 k)^{2}}{k}-k=k .
\end{array}
$$

Summing over all vertices, the number of isosceles triangles is at least
$$
\sum_{v \in V} \sum_{i=1}^{k} \mathrm{C}_{d(v)}^{2} \geqslant \sum_{v \in V} k=k(2 k+1) .
$$

Since there are no monochromatic triangles, each triangle is either an isosceles triangle or a triangle with three edges of different colors (let the number of such triangles be $A$). Therefore,
$$
A \leqslant \mathrm{C}_{2 k+1}^{3}-k(2 k+1)=\frac{2 k(k-2)(2 k+1)}{3} \text {. }
$$

Below is an example where $A=\frac{2 k(k-2)(2 k+1)}{3}$.
Considering the condition for equality in the above inequality is that for all $i, v$, we have $d_{i}(v)=2$.

Thus, it suffices to prove that for any complete graph with $2 k+1$ vertices, it can be divided into $k$ pairwise disjoint Hamiltonian cycles.

This can be achieved by coloring each Hamiltonian cycle with a different color.

In fact, starting from vertex $u_{0}$ and alternately connecting two sequences of vertices $v_{1}, v_{2}, \cdots, v_{k}$ and $v_{2 k}, v_{2 k-1}, \cdots, v_{k+1}$, we obtain a cycle
$$
C_{0}=\left(u_{0}, v_{1}, v_{2 k}, v_{2}, v_{2 k-1}, \cdots, v_{k}, v_{k+1}, u_{0}\right) .
$$

By adding $i$ to the index of each vertex in cycle $C_{0}$ (excluding $u_{0}$), we obtain cycle $C_{i}(1 \leqslant i \leqslant k-1)$, where for any integer $j$, $v_{j}=v_{j+2 k}$.
It can be verified that these $k$ Hamiltonian cycles are pairwise disjoint.