$4 \cdot 35$ Find the smallest positive integer $n$, such that in any two-coloring of the complete graph $K$ with $n$ vertices, there always exist $m$ edges of the same color, which are pairwise non-adjacent.

 untranslated parts:
求最小正整数 $n$, 使在任何二染色的 $n$ 个顶点的完全图 $K$,中,总存在同色的 $m$ 条线段,它们两两之间没有公共端点。

translated:
Find the smallest positive integer $n$, such that in any two-coloring of the complete graph $K$ with $n$ vertices, there always exist $m$ edges of the same color, which are pairwise non-adjacent.
[Solution] Clearly, when $m=1$, $n=2$. When $m=2$, $n=5$. In fact, in the two-colored $K_{4}$ (solid lines represent blue, dashed lines represent red) shown in the figure, any two segments of the same color have a common endpoint. However, in a one-colored $K_{5}$, if all segments are the same color, there are certainly two segments of the same color without a common point; otherwise, there must be one red and one blue segment with one common endpoint, say $A_{1} A_{2}$ is red, and $A_{2} A_{3}$ is blue. It is evident that, regardless of the color of $A_{4} A_{5}$, there are always 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 natural numbers $m$, $n=3 m-1$.
We have already proven the proposition for $m=1,2$. Assume the proposition holds for $m=k$. When $m=k+1$, if all edges of $K_{3 k+2}$ are the same color, it is certainly possible to find $k+1$ segments that meet the requirement. Otherwise, there must be one vertex that is the common endpoint of one red edge and one blue edge. Removing these three endpoints, there are still $3 k-1$ vertices left. By the induction hypothesis, in the complete subgraph $K_{3 k-1}$ with these $3 k-1$ vertices, there exist $k$ segments of the same color, none of which have a common endpoint. Adding one more segment of the same color from the previous two, we get $k+1$ segments, all of the same color and none of which have a common endpoint.

On the other hand, divide the vertices of $K_{3 k+1}$ into two groups: Group $A$ has $k$ vertices, and Group $B$ has $2 k+1$ vertices. Color all the segments between the vertices in Group $B$ red, and all other segments blue. Thus, whether it is $k+1$ red segments or $k+1$ blue segments, there are always two segments with one common endpoint. This shows that the smallest natural number $n=3 k+2$, completing the induction proof.