$4 \cdot 36$ Find the smallest positive integer $n$, such that in any two-coloring of $K_{n}$, there exist 3 monochromatic triangles, each pair of which has no common edge.

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[Solution] In the given two-colored $K_{8}$, there are two blue $K_{4}$, and the rest of the unshown edges are all red. It is easy to see that there are no red triangles and exactly 8 blue triangles in the graph. However, among any 3 blue triangles, there are always 2 triangles in the same blue $K_{4}$, sharing 1 common edge. Therefore, the smallest positive integer $n \geqslant 9$.

For a two-colored $K_{9}$, let $\triangle A_{1} A_{2} A_{3}$ be one of its monochromatic triangles. Consider the complete subgraph $K_{6}$ with vertices $\left\{A_{4}, A_{5}, A_{6}, A_{7}, A_{8}, A_{9}\right\}$, which must contain a monochromatic triangle. Suppose $\triangle A_{7} A_{8} A_{9}$ is a blue triangle. Now consider the $K_{5}$ with vertices $\left\{A_{3}, A_{4}, A_{5}, A_{6}, A_{7}\right\}$. If it contains a monochromatic triangle, then together with the previous two, there are 3 monochromatic triangles with no common edges. If it does not contain a monochromatic triangle, it can be decomposed into one red and one blue cycle, each with 5 edges (see figure). Similarly, in the $K_{5}$ with vertices $\left\{A_{3}, A_{4}, A_{5}, A_{6}, A_{8}\right\}$ and $\left\{A_{3}, A_{4}, A_{5}, A_{6}, A_{9}\right\}$, if either $K_{5}$ contains a monochromatic triangle, the problem is solved. Otherwise, both $K_{5}$ can be decomposed into one red and one blue cycle, each with 5 edges. Since the colors of the $K_{4}$ with vertices $\left\{A_{3}, A_{4}, A_{5}, A_{6}\right\}$ are already determined, it must be that $A_{3} A_{8}, A_{3} A_{9}, A_{6} A_{8}, A_{6} A_{9}$ are all blue edges. Thus, $\triangle A_{3} A_{7} A_{9}$ and $\triangle A_{6} A_{7} A_{8}$ are both blue triangles. Adding the initial monochromatic triangle $A_{1} A_{2} A_{3}$ meets the requirements of the problem.
In summary, the smallest positive integer $n=9$.