4.41 Find the smallest positive integer $n$, such that in any two-coloring of $K_{n}$, there exist $m$ pairwise vertex-disjoint monochromatic triangles.
[Solution] Clearly, when $m=1$, $n=6$. Now let's examine when $m=2$, what is the smallest positive integer $n$. At this point, in the two-coloring of $K_{7}$ shown in the figure, only the blue edges are drawn, and the rest are red. It is easy to see that there are no red triangles, and the 10 blue triangles are all within the blue $K_{5}$ below, and of course, any two have a common vertex. From this, we can conclude that when $m=2$, the smallest positive integer $n \geqslant 8$.

For a two-coloring of $K_{8}$, let's assume $\triangle A_{1} A_{2} A_{3}$ is one of the blue triangles. If the $K_{5}$ with the other 5 points $\left\{B_{1}, B_{2}, B_{3}, B_{4}, B_{5}\right\}$ as vertices also has a monochromatic triangle, the problem is solved, so we only need to discuss the case where there are no monochromatic triangles. At this point, the two-coloring of $K_{5}$ can be decomposed into one red and one blue circle, each with 5 edges (as shown in the figure). In addition, in the two-coloring $K_{6}$ with vertices $\left\{A_{i}, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}\right\}$, there must be two monochromatic triangles with $A_{i}$ as a common vertex, $i=1,2,3$. In the 3 $K_{6}$s, there are a total of 6 monochromatic triangles.
(1) Suppose there is 1 red triangle, let's assume it is $\triangle A_{1} B_{1} B_{4}$. Consider the $K_{5}$ with vertices $\left\{A_{2}, A_{3}, B_{2}, B_{3}, B_{5}\right\}$. If there is a monochromatic triangle in it, then $\triangle A_{1} B_{1} B_{4}$ and this triangle form a pair that meets the requirement. Otherwise, in this two-coloring $K_{5}$, each vertex has two red edges and two blue edges. But $B_{5} B_{2}$ and $B_{5} B_{3}$ are already red edges, so $B_{5} A_{2}$ and $B_{5} A_{3}$ should be blue edges, leading to $\triangle B_{5} A_{2} A_{3}$ being a blue triangle, which is a contradiction.
(2) Suppose all 6 monochromatic triangles are blue, then in each $K_{6}$, the two blue triangles share 1 edge. Let the two blue triangles in the first $K_{6}$ be $\triangle A_{1} B_{1} B_{2}$ and $\triangle A_{1} B_{2} B_{3}$, i.e., $B_{2}$ is the common vertex of the two blue triangles. If one of the pairs of blue triangles in the next two $K_{6}$s does not have $B_{2}$ as a common vertex, then there must be two blue triangles with no common vertex; if all three pairs also have $B_{2}$ as a common vertex, i.e., $\triangle A_{2} B_{1} B_{2}$, $\triangle A_{2} B_{2} B_{3}$ and $\triangle A_{3} B_{1} B_{2}$, $\triangle A_{3} B_{2} B_{3}$ are all blue triangles. In this case, $\triangle A_{1} B_{1} B_{2}$ and $\triangle A_{2} A_{3} B_{3}$ are two blue triangles with no common vertex.

Thus, we have proven that when $m=2$, the smallest positive integer $n=3 m+2=8$. For general $m$, we use induction to prove that the same conclusion holds.

Assume that when $m=k$, the smallest positive integer $n=3 m+2$. When $m=k+1$, in the two-coloring of $K_{3 k+5}$, there is certainly a monochromatic triangle. Apart from the 3 vertices of this triangle, there are still $3 k+2$ vertices. By the induction hypothesis, in the $K_{3 k+2}$ with these $3 k+2$ vertices, there are $k$ monochromatic triangles with no common vertices. Adding the initial 1, there are a total of $k+1$ monochromatic triangles with no common vertices.

On the other hand, as at the beginning, take two vertices $A_{1}$ and $A_{2}$ from the $3 m+1$ vertices and color $A_{1} A_{2}$ blue. Then color the complete subgraph $K_{3 m-1}$ with the remaining vertices all blue, and the rest of the edges red. Thus, this $K_{3 m+1}$ has no red triangles, and the blue triangles are all in the subgraph $K_{3 m-1}$, so any $m$ monochromatic triangles must have at least 1 common vertex. That is, this $K_{3 m+1}$ does not have $m$ monochromatic triangles with no common vertices. Therefore, when $m \geqslant 2$, the smallest positive integer $n=3 m+2$; when $m=1$, $n=6$.