Example 3 A club has 99 people. Find the smallest positive integer $n$ such that if each person knows more than $n$ people, there must exist 4 people who all know each other (here it is agreed that if $A$ knows $B$, then $B$ also knows $A$).
We use 99 points to represent 99 people. If two people do not know each other, then a line is drawn between the corresponding points, thus forming a graph $G$. For each point $v$ in $G$, we have $d(v) \leqslant (99-1)-(n+1) = 97-n$. Take any point $A$, then $d(A) \leqslant 97-n$. If $97-n < \frac{97}{2}$, there must exist a point $B$ not adjacent to $A$; if $d(A) + d(B) = 2(97-n) < 97$, there must exist a point $C$ not adjacent to both $A$ and $B$; finally, if $d(A) + d(B) + d(C) = 3(97-n) < 65$, there must exist a point $D$ not adjacent to $A, B, C$, i.e., when $n \geqslant 66$, there exist 4 points $A, B, C, D$ in the graph that are pairwise non-adjacent, meaning the 4 corresponding people all know each other.

On the other hand, if each person knows no more than 66 people, we can take 3 complete graphs $K_{33}$ to form a 99-vertex graph $G$, where each point is not adjacent to exactly 66 points, i.e., each person has exactly 66 acquaintances. In this case, among any 4 people, the corresponding 4 points must have 2 points belonging to the same $K_{33}$, and the 2 corresponding people do not know each other. Therefore, it is impossible for 4 people to all know each other.
In summary, the minimum value of $n$ is 66.