Example 7 Given that in any 3 people among $n$ people, there must be 2 who know each other, if there must be 4 people who all know each other, find the minimum value of $n$.


Solve: Use $n$ points to represent $n$ people. If two people know each other, the line connecting the corresponding points is colored red; if two people do not know each other, the line connecting the corresponding points is colored blue. Thus, the problem is transformed into finding the smallest positive integer $n$ such that if a 2-colored complete graph contains no blue triangle, then it must contain a red complete graph $K_{4}$.

First, as shown in Figure 7-7, in a 2-colored graph $K_{8}$, there is neither a blue triangle (represented by thick solid lines) nor a red complete graph $K_{4}$ (represented by thin dashed lines). Therefore, the smallest positive integer $n \geqslant 9$.

Second, assume that in a 2-colored $K_{9}$, there is no blue triangle. Consider the 8 edges starting from a certain vertex, divided into the following three cases:
(1) If there exists a vertex $A_{1}$, from which at least 4 blue edges start, without loss of generality, let $A_{1} A_{2}, A_{1} A_{3}, A_{1} A_{4}, A_{1} A_{5}$ be blue. Since there is no blue triangle, the $K_{4}$ with vertices $A_{2}, A_{3}, A_{4}, A_{5}$ is a red complete graph.
(2) If there exists a vertex $A_{1}$, from which at most 2 blue edges start, thus at least 6 red edges start, without loss of generality, let $A_{1} A_{2}, A_{1} A_{3}, A_{1} A_{4}, A_{1} A_{5}, A_{1} A_{6}, A_{1} A_{7}$ be red. Consider the 2-colored complete graph $K_{6}$ with vertices $A_{2}, A_{3}, \cdots, A_{7}$. By Ramsey's theorem, it must contain a monochromatic triangle. Since there is no blue triangle, there must be a red triangle. Without loss of generality, let $\triangle A_{2} A_{3} A_{4}$ be a red triangle. Thus, the $K_{4}$ with vertices $A_{1}, A_{2}, A_{3}, A_{4}$ is a red complete graph.
(3) If from each vertex exactly 3 blue edges start, then the number of blue edges in the graph is $\frac{1}{2} \times 9 \times 3=13 \frac{1}{2}$, which is not an integer. Therefore, this case is impossible.
This proves that in a 2-colored $K_{9}$, if there is no blue triangle, then there must be a red $K_{4}$.
In summary, the smallest positive integer $n$ is 9.