Example 18 MO Space City consists of 99 space stations, with tubular passages connecting any two stations. It is stipulated that 99 of these passages are bidirectional main roads, while the rest are strictly one-way passages. If any four space stations can reach any other station among them through the passages, then the set of these four stations is called a fully connected quartet.

Try to design a scheme for MO Space City to maximize the number of fully connected quartets (please calculate the exact maximum number and prove your conclusion).
(1999 CMO Problem)
Solve by generalizing the problem. Below, we discuss the general case of $n$ space stations and $n$ bidirectional main lines, where $n$ is an odd number greater than 3, and let $m=\frac{1}{2}(n-3)$. In this problem, $n=99, m=48$.
(1) If among four space stations, one station has channels that are strictly unidirectional from that station to the other three, then this set of four stations is not a mutually connected four-station group. We classify such non-mutually connected four-station groups into class $S$, and the rest of the non-mutually connected four-station groups into class $T$. Thus, the total number of mutually connected four-station groups is
$$
N_{n}=C_{n}^{4}-|T|-|S| .
$$

Label the $n$ space stations as $1,2, \cdots, n$. Let the number of strictly unidirectional channels from station $i$ be $x_{i}$. Then the total number of non-mutually connected four-station groups in class $S$ is $|S|=\sum_{i=1}^{n} C_{x_{i}}^{3}$, and
$$
x_{1}+x_{2}+\cdots+x_{n}=C_{n}^{2}-n=\frac{1}{2} n(n-3)=n m .
$$

To maximize $N_{n}$, $|T|$ and $|S|$ must be minimized.
First, we prove that $|S| \geqslant n C_{m}^{3}$. In fact, when $|S|$ takes its minimum value, for any $1 \leqslant i < j \leqslant n$, if $x_{i} < m$ and $x_{j} > m$, then by increasing $x_{i}$ by 1 and decreasing $x_{j}$ by 1, the value of $|S|$ decreases (since $\left(x_{j}-1\right)-x_{i} \geqslant 1$), which contradicts the minimality of $|S|$.
Thus, $|S| \geqslant n C_{m}^{3}$, and equality holds if and only if $x_{1}=x_{2}=\cdots=x_{n}=m$. Therefore,
$$
N_{n}=C_{n}^{4}-|T|-|S| \leqslant C_{n}^{4}-0-n C_{m}^{3}=\frac{1}{48} n(n-3)\left(n^{2}+6 n-31\right).
$$
(2) The following design shows that $N_{n}=C_{n}^{4}-n C_{m}^{3}$ can be achieved.
First, arrange the space stations numbered $1,2, \cdots, n$ in a clockwise order on $n$ points $A_{1}$, $A_{2}, \cdots, A_{n}$ on a circle. The channels between adjacent space stations on the circle are bidirectional main lines, thus setting up $n$ bidirectional main lines:
$A_{1} A_{2}, A_{2} A_{3}, \cdots, A_{n-1} A_{n}, A_{n} A_{1}$.
For any $i, j \in\{1,2,3, \cdots, n\}, i \neq j$, if the arc from $A_{i}$ to $A_{j}$ in the clockwise direction passes through an odd number of space stations, then the channel between $A_{i}$ and $A_{j}$ is a strictly unidirectional channel from $A_{i}$ to $A_{j}$: $A_{i} \rightarrow A_{j}$. Since $n$ is odd, one of the clockwise arcs from $A_{i}$ to $A_{j}$ and from $A_{j}$ to $A_{i}$ passes through an odd number of space stations. Hence, the above rule does not lead to any contradiction.

According to this design, the number of strictly unidirectional channels from each $A_{i}$ is $m=\frac{n-3}{2}$, so $|S|=$ $n C_{m}^{3}$. We now prove that in this design, $|T|=0$.

If a four-station group has a bidirectional main line between two stations $A$ and $B$, then it is easy to see that this four-station group is mutually connected. Therefore, if a four-station group $A, B, C, D$ is not mutually connected, then there are no bidirectional main lines between any two stations. Let the number of space stations between $A$ and $B$, $B$ and $C$, $C$ and $D$, and $D$ and $A$ be $a$, $b$, $c$, and $d$, respectively. Thus, $a+b+c+d=n-4$ is odd. Therefore, the number of odd numbers among $a, b, c, d$ is 1 or 3.
(i) If $a$ is odd and $b, c, d$ are even, then $A \rightarrow B \rightarrow D \rightarrow C \rightarrow A$, and $A, B, C, D$ form a mutually connected four-station group (Figure 4-15(a)).
(ii) If $a, b, c$ are odd and $d$ is even, then the channels from $A$ to $B, C, D$ are all strictly unidirectional channels from $A$. Such a non-mutually connected four-station group belongs to class $S$ (Figure 4-15(b)).
From the above discussion, we know that in this design, $|T|=0$.
In summary, the maximum number of mutually connected four-station groups is $C_{n}^{4}-n C_{\frac{3-3}{2}}^{3}=\frac{1}{48} n(n-3)\left(n^{2}+6 n-31\right)$. Specifically, when $n=99$, the maximum number of mutually connected four-station groups in this problem is $C_{99}^{4}-99 C_{48}^{3}=2052072$.