Four. (50 points) There are 2008 students participating in a large public welfare activity. If two students know each other, then these two students are considered as a cooperative group.
(1) Find the minimum number of cooperative groups $m$, such that no matter how the students know each other, there exist three students who are pairwise in a cooperative group;
(2) If the number of cooperative groups is $\left[\frac{m}{22}\right]$, prove: there exist four students $A, B, C, D$ such that $A$ and $B$, $B$ and $C$, $C$ and $D$, $D$ and $A$ are each in a cooperative group.
(1) Let $n=1004$. Below is the proof: $m=n^{2}+1$. Divide the students into two large groups, each with $n$ students, and no student in the same large group knows each other, while each student knows every student in the other large group, thus forming $n^{2}$ cooperative groups, but there do not exist three students who are all in a cooperative group with each other.
If there are $n^{2}+1$ cooperative groups, let student $X$ know the most students, and let these students be $Y_{1}, Y_{2}, \cdots, Y_{k}$.
If no two students among $Y_{1}, Y_{2}, \cdots, Y_{k}$ are in a cooperative group, then the number of cooperative groups does not exceed
$k+(2 n-k-1) k=k(2 n-k) \leqslant n^{2}$, leading to a contradiction. Therefore, $m=1004^{2}+1=1008017$.
(2) Let $n=1004, l=\left[\frac{m}{22}\right]=45818, 2 n$ students are $X_{1}, X_{2}, \cdots, X_{2 n}$, and the number of students they know are $a_{1}, a_{2}, \cdots, a_{2 n}$, respectively, then $a_{1}+a_{2}+\cdots+a_{2 n}=2 l$. Consider all possible pairs of students among the students each student knows, the total number is
$$
\begin{aligned}
C_{a_{1}}^{2}+ & C_{a_{2}}^{2}+\cdots+C_{a_{n}}^{2}=\frac{1}{2}\left(\sum_{i=1}^{2 n} a_{i}^{2}-\sum_{i=1}^{2 n} a_{i}\right) \geqslant \frac{1}{2}\left[\frac{1}{2 n}\left(\sum_{i=1}^{2 n} a_{i}\right)^{2}-\sum_{i=1}^{2 n} a_{i}\right] \\
& =\frac{1}{2} \sum_{i=1}^{2 n} a_{i}\left(\frac{1}{2 n} \sum_{i=1}^{2 n} a_{i}-1\right)=\frac{2 l}{2}\left(\frac{2 l}{2 n}-1\right)=l\left(\frac{l}{n}-1\right)=45818\left(\frac{45818}{1004}-1\right)>C_{2008}^{2} .
\end{aligned}
$$

Therefore, there exists a pair of students $A$ and $C$, who both know $B$ and $D$.