19. Find the smallest positive integer $n$, such that in any 2-coloring of $K_{n}$, there exist two monochromatic triangles of the same color with no common edges.
(1991 Chinese National Training Team Problem)
19. As shown in Figure 1, in a 2-colored $K_{7}$, solid lines represent blue, and dashed lines represent red. Among them, any two red triangles and any two blue triangles share a common edge, indicating that the required smallest positive integer $n \geqslant 8$. Next, we only need to prove that in a 2-colored $K_{8}$, there must exist two triangles of the same color that do not share a common edge. If this conclusion does not hold, then there exists a 2-colored $K_{8}$ in which there are no two triangles that meet the requirements of the problem. By Ramsey's theorem, a 2-colored $K_{8}$ must contain a monochromatic triangle. Without loss of generality, let $\triangle A_{1} A_{2} A_{3}$ be a blue triangle. Consider the 2-colored complete subgraph $K_{6}$ with vertices $\left\{A_{3}, A_{4}, A_{5}, A_{6}, A_{7}, A_{8}\right\}$. By Ramsey's theorem, it must contain a monochromatic triangle. By the assumption of contradiction, it cannot contain another blue triangle, so it must be a red triangle. This red triangle must share a common vertex with $\triangle A_{1} A_{2} A_{3}$, otherwise, among the 9 edges connecting the vertices of the two triangles, there must be 5 edges of the same color. Suppose these are 5 red edges. Then, by the pigeonhole principle, one of the vertices $A_{1}, A_{2}, A_{3}$ must connect to the red triangle with at least $\left[\frac{5-1}{3}\right]+1=2$ red edges, thus forming a red triangle that shares a vertex with the blue $\triangle A_{1} A_{2} A_{3}$. Let this triangle be $\triangle A_{3} A_{4} A_{5}$. Consider the 2-colored $K_{5}$ with vertices $\left\{A_{2}, A_{4}, A_{6}, A_{7}, A_{8}\right\}$. By the assumption of contradiction, it does not contain a monochromatic triangle, so it can be decomposed into a red and a blue cycle, each with 5 edges (as shown in Figure 2). Consider the 2-colored $K_{6}$ with vertices $\left\{A_{1}, A_{2}, A_{4}, A_{6}, A_{7}, A_{8}\right\}$. It must contain 2 monochromatic triangles (Example 1 of $83-1$). By the assumption of contradiction, it cannot contain a red triangle, so it must contain 2 blue triangles, both of which share an edge with the blue $\triangle A_{1} A_{2} A_{3}$. Of course, this edge can only be $A_{1} A_{2}$, so $A_{1} A_{4}$ and $A_{1} A_{6}$ must be blue edges. Consider the 2-colored $K_{6}$ with vertices $\left\{A_{2}, A_{4}, A_{5}, A_{6}, A_{7}, A_{8}\right\}$. Similarly, we can conclude that $A_{5} A_{6}$ and $A_{4} A_{6}$ are red edges. Finally, consider $A_{3} A_{6}$. If it is a blue edge, then $\triangle A_{2} A_{3} A_{6}$ and $\triangle A_{2} A_{1} A_{4}$ are two blue triangles that do not share a common edge, which is a contradiction; if $A_{3} A_{6}$ is a red edge, then $\triangle A_{3} A_{4} A_{6}$ and $\triangle A_{4} A_{5} A_{7}$ are two red triangles that do not share a common edge, which is also a contradiction. This proves that in a 2-colored $K_{8}$, there must exist two triangles of the same color that do not share a common edge. In summary, the required smallest positive integer $n=8$.