Example 10 Find the smallest positive integer $n$, such that in any two-coloring of the complete graph $K_{n}$ with $n$ vertices, there always exist $m$ monochromatic edges, no two of which share a common endpoint.
Solution: It is clear that when $m=1$, $n=2$; when $m=2$, $n=5$. In fact, in the 2-colored $K_{4}$ as shown in Figure 7-11, any two segments of the same color have a common endpoint (solid lines represent red, dashed lines represent blue). However, in a 2-colored $K_{5}$, if all segments are the same color, there are certainly two segments of the same color without a common endpoint. Otherwise, there must be a red segment and a blue segment with a common endpoint. Let's assume $A_{1} A_{2}$ is red and $A_{2} A_{3}$ is blue. Clearly, regardless of the color of $A_{4} A_{5}$, there will be two segments of the same color without a common endpoint. This proves that when $m=2$, the smallest positive integer $n=5$.

Next, we use induction to prove that for all positive integers $m$, $n=3m-1$.

The proposition has been proven for $m=1,2$. Assume the proposition holds for $m=r$. When $m=r+1$, if all edges of $K_{3r+2}$ are the same color, there are certainly $r+1$ segments of the same color, none of which have a common endpoint; otherwise, there must be a vertex that is the common endpoint of a red edge and a blue edge. Removing the 3 endpoints of these two edges leaves $3r-1$ vertices. By the induction hypothesis, in the $K_{3r-1}$ with these $3r-1$ vertices, there exist $r$ segments of the same color, none of which have a common endpoint. Adding one of the two original segments of the same color, we obtain $r+1$ segments of the same color, none of which have a common endpoint.

On the other hand, divide the vertices of $K_{3r+1}$ into two groups: Group $A$ has $r$ vertices, and Group $B$ has $2r+1$ vertices. Color the segments between the vertices of Group $B$ red, and all other segments blue. Thus, whether the $r+1$ segments are red or blue, there will always be two segments with a common endpoint, indicating that the smallest positive integer $n=3r+2$. Therefore, we have completed the induction proof.