7・120 MO Space City consists of 99 space stations. Any two space stations are connected by a tube-shaped passage. It is stipulated that 99 of these passages are bidirectional main roads, and the rest are strictly one-way passages. If any 4 space stations can be reached from any other station among them via the passages, then this set of 4 stations is called an interconnected quad station group. Try to design a scheme for the $M O$ Space City to maximize the number of interconnected quad station groups, calculate this maximum number, and prove your conclusion.
[Solution] We refer to a group of four stations that cannot communicate with each other as a bad four-station group. Thus, there are 3 possible scenarios for a bad four-station group:
(1) Station $A$ has 3 outgoing channels $A B, A C, A D$ all leaving $A$;
(2) Station $A$ has 3 incoming channels all entering $A$;
(3) Stations $A$ and $B$, and stations $C$ and $D$ are all connected by two-way roads, but channels $A C, A D$ all leave $A$, and channels $B C, B D$ all leave $B$.

Let the set of all bad four-station groups of the first type be denoted as $S$, and the set of other bad four-station groups be denoted as $T$. Let's calculate $|S|$. Note that in the space city, there are
$$
C_{99}^{2}-99=\frac{1}{2} \times 99 \times 98-99=99 \times 48
$$

single-direction channels. Let the number of outgoing channels from station $i$ be $S_{i}$. Therefore,
$$
\sum_{i=1}^{99} S_{i}=99 \times 48 \text {. }
$$

The number of type 1 bad four-station groups with 3 outgoing channels from station $A_{i}$ is $C_{S_{i}}^{3}$. Thus,
$$
|S|=\sum_{i=1}^{99} C_{S_{i}}^{3} \geqslant 99 \times C_{48}^{3}=99 \times 8 \times 47 \times 46 .
$$

Since the total number of four-station groups is
$$
C_{99}^{4}=4 \times 99 \times 98 \times 97,
$$

the number of interconnected four-station groups is no more than
$$
\begin{aligned}
C_{99}^{4}-|S| & \leqslant 4 \times 99 \times 98 \times 97-8 \times 99 \times 47 \times 46 \\
& =8 \times 99(49 \times 97-47 \times 46) \\
& =2052072 .
\end{aligned}
$$

Next, we construct an example where the number of interconnected four-station groups is exactly 2052072. For this, the number of outgoing channels from each station $A_{i}$ must be 48. Thus, the number of incoming channels is also 48, and each station has exactly two two-way channels, while ensuring that there are only $S$ type bad four-station groups and no $T$ type bad four-station groups.

Write the 99 space stations on the 99 vertices of a regular 99-sided polygon inscribed in a circle. Define the longest diagonal of the regular 99-sided polygon as a two-way channel. Thus, each point has exactly two two-way channels. For station $A_{i}$, the 48 stations in the clockwise direction from $A_{i}$ are connected to $A_{i}$ by single-direction channels all leaving $A_{i}$; the 48 stations in the counterclockwise direction are connected to $A_{i}$ by single-direction channels all entering $A_{i}$. Below, we verify that under this definition, there are only $S$ type bad four-station groups and no $T$ type bad four-station groups.
Let $\{A, B, C, D\}$ be any four-station group.
(i) If there are two two-way channels among the 4 stations, they are obviously interconnected;
(ii) If there is only one two-way channel $A C$ among the 4 stations, then stations $B$ and $D$ form a loop with $A$ and $C$, respectively, and are thus interconnected.
(iii) From (i) and (ii), we know that bad four-station groups have no two-way channels, so they can only be one of the two scenarios (1) and (2). If it is (2), assume that the 3 channels of station $A$ are all entering $A$. Then stations $B, C, D$ are all within the 48 stations in the counterclockwise direction from $A$. Let station $D$ be the farthest from $A$, so channels $A D, B D, C D$ all leave station $D$. This indicates that all bad four-station groups are of type $S$.
In summary, the maximum number of interconnected four-station groups is 2052072.