$4 \cdot 42$ Find the smallest positive integer $n$, such that in any two-coloring of the complete graph $K_{n}$ with $n$ vertices, there exist two monochromatic triangles of the same color that have no common vertices.
[Solution] First, let's look at the right figure showing a two-coloring of $K_{9}$, where the dashed lines represent red, the solid lines represent blue, and the lines not drawn are all dashed. It is easy to see that the blue $K_{5}$ with vertices $\left\{B_{1}, B_{2}, B_{3}, B_{4}, B_{5}\right\}$ contains 10 blue triangles, any two of which share at least 1 vertex. There are no other blue triangles in the graph. Moreover, any red triangle in the graph must have one edge among the three segments $\left\{A_{2} A_{3}, A_{3} A_{4}, A_{4} A_{2}\right\}$. Therefore, any two red triangles also share at least 1 vertex. This shows that the smallest positive integer $n \geqslant 10$.

Next, we will prove that in a two-coloring of $K_{10}$, there always exist two monochromatic triangles that satisfy the requirements of the problem. Assume this conclusion is not true.
(1) Since a monochromatic triangle must exist in a two-coloring of $K_{6}$, it certainly exists in $K_{10}$. Let $\triangle A_{1} A_{2} A_{3}$ be a monochromatic triangle. Then, in the $K_{7}$ with vertices $\left\{A_{4}, A_{5}, \cdots, A_{10}\right\}$, there must be another monochromatic triangle, which we can assume to be $\triangle A_{4} A_{5} A_{6}$. These two monochromatic triangles have no common vertices, so they must be one red and one blue. Assume $\triangle A_{1} A_{2} A_{3}$ is red and $\triangle A_{4} A_{5} A_{6}$ is blue. Consider the 9 lines connecting the vertices of these two triangles. By the pigeonhole principle, at least 5 of these lines must be the same color, let's say blue. These 5 blue lines are drawn from $A_{1}, A_{2}, A_{3}$, so at least one vertex must have two blue lines, say $A_{1} A_{4}, A_{1} A_{5}$. Thus, $\triangle A_{1} A_{4} A_{5}$ is a blue triangle, which shares exactly 1 vertex $A_{1}$ with the red $\triangle A_{1} A_{2} A_{3}$. This means that, by the assumption of contradiction, the two-colored complete subgraph $K_{5}$ with vertices $\left\{B_{1}, B_{2}, B_{3}, B_{4}, B_{5}\right\}$ has no monochromatic triangles, so this $K_{5}$ must be decomposed into one red and one blue cycle, each with exactly 5 edges. Consider the complete subgraph $K_{6}$ with vertices $\left\{A_{1}, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}\right\}$. At this point, there must be two monochromatic triangles. Clearly, both share the vertex $A_{1}$.
(2) Suppose these two monochromatic triangles are the same color, let's say blue, then they must share one edge. Let these two blue triangles be $\triangle A_{1} B_{1} B_{5}$ and $\triangle A_{1} B_{5} B_{4}$ (see figure). Consider the complete subgraph $K_{5}$ with vertices $\left\{B_{2}, B_{3}, B_{4}, A_{4}, A_{5}\right\}$. By the assumption of contradiction, it has no monochromatic triangles, so it can also be decomposed into one red and one blue cycle, each with exactly 5 edges. Assume $A_{4} B_{2}, A_{5} B_{4}$ are blue edges, then $A_{4} B_{3}, A_{4} B_{4}, A_{5} B_{2}, A_{5} B_{3}$ are red edges. Similarly, the complete subgraph $K_{5}$ with vertices $\left\{B_{1}, B_{2}, B_{3}, A_{4}, A_{5}\right\}$ can also be decomposed into one red and one blue cycle, each with exactly 5 edges. Therefore, vertex $B_{3}$ must draw two red lines in this $K_{5}$. But it has already been proven that $B_{3} B_{1}, B_{3} A_{4}, B_{3} A_{5}$ are all red edges, which is a contradiction.
(3) From (2), we know that the two monochromatic triangles must be different colors. Assume $\triangle A_{1} B_{2} B_{3}$ is blue and $\triangle A_{1} B_{1} B_{4}$ is red. Consider the complete subgraph $K_{5}$ with vertices $\left\{A_{4}, A_{5}, B_{1}, B_{4}, B_{5}\right\}$. By the assumption of contradiction, it has no monochromatic triangles, so it can be decomposed into one red and one blue cycle, each with exactly 5 edges. Assume $A_{5} B_{1}, A_{4} B_{4}$ are blue edges and $A_{4} B_{1}, A_{4} B_{5}, A_{5} B_{4}, A_{5} B_{5}$ are red edges (as shown in the figure). If $A_{5} B_{2}$ is a red edge, then $\triangle A_{5} B_{2} B_{4}$ and $\triangle A_{1} A_{2} A_{3}$ are two red triangles with no common vertices, which contradicts the assumption of contradiction, so $A_{5} B_{2}$ is a blue edge. If $A_{4} B_{3}$ is a red edge, then $\triangle A_{4} B_{3} B_{5}$ and $\triangle A_{1} A_{2} A_{3}$ are two red triangles with no common vertices, which is a contradiction, so $A_{4} B_{3}$ is a blue edge. At this point, $\triangle A_{4} B_{3} B_{4}$ and $\triangle A_{1} B_{2} A_{5}$ are two blue triangles with no common vertices, which contradicts the assumption of contradiction.

In summary, we have proven that in a two-coloring of $K_{10}$, there always exist two monochromatic triangles that satisfy the requirements. Therefore, the smallest positive integer $n=10$.