$4 \cdot 43$ Find the smallest natural number $n$, such that in any two-coloring of $K_{n}$, there always exist two monochromatic triangles, each sharing exactly 1 vertex.
[Solution] Color the 8-vertex complete graph $K_{8}$ as shown in the figure, where blue lines are drawn, and all undrawn lines are red. It is easy to see that there are 8 blue triangles and no red triangles in the graph, and any two blue triangles either share a common edge or have no common vertex. Therefore, the smallest natural number $n \geqslant 9$.

To prove that in a two-colored $K_{9}$, there must be two monochromatic triangles with exactly 1 common vertex, we first prove the following lemma.
Lemma: In a two-colored $K_{9}$, there are at least 12 different monochromatic triangles.
Proof of the lemma: Let the 9 vertices of $K_{9}$ be $A_{1}, A_{2}, \cdots, A_{9}$, where $r_{i}$ red edges and $8-r_{i}$ blue edges are drawn from vertex $A_{i}$. We call the angle formed by a red edge and a blue edge from a vertex an "heterochromatic angle." Thus, the number of heterochromatic angles with $A_{i}$ as the vertex is $r_{i}\left(8-r_{i}\right)$. We call a triangle that has both red and blue edges a "heterochromatic triangle." Clearly, each heterochromatic triangle has exactly two heterochromatic angles. Therefore, the number of heterochromatic triangles in a two-colored $K_{9}$ is
$$
\frac{1}{2} \sum_{i=1}^{9} r_{i}\left(8-r_{i}\right) \leqslant \frac{1}{2} \times 9 \times 16=72,
$$

i.e., the number of heterochromatic triangles does not exceed 72. Since the total number of triangles in $K_{9}$ is $C_{9}^{3}=\frac{1}{6} \times 9 \times 8 \times 7=84$, there are at least 12 monochromatic triangles in $K_{9}$. The lemma is proved.

By the lemma, there are at least 12 monochromatic triangles in a two-colored $K_{9}$. These 12 triangles have 36 vertices in total. By the pigeonhole principle, there must be 1 vertex that is the common vertex of 4 monochromatic triangles. If any two of these monochromatic triangles do not have exactly 1 common vertex, then they must share a common edge. Therefore, the 4 monochromatic triangles share 1 common edge and are of the same color. Assume $\triangle A_{1} A_{2} A_{3}, \triangle A_{1} A_{2} A_{4}, \triangle A_{1} A_{2} A_{5}, \triangle A_{1} A_{2} A_{6}$ are all blue triangles. Consider the two-colored $K_{4}$ with vertices $A_{3}, A_{4}, A_{5}, A_{6}$. If there is 1 blue edge, say $A_{3} A_{4}$, then $\triangle A_{1} A_{3} A_{4}$ and $\triangle A_{1} A_{2} A_{5}$ are both blue triangles and have exactly 1 common vertex $A_{1}$; if all edges are red, then red $\triangle A_{3} A_{4} A_{5}$ and blue $\triangle A_{1} A_{2} A_{3}$ have exactly 1 common vertex $A_{3}$. This proves that in a two-colored $K_{9}$, there must be two monochromatic triangles that meet the requirements of the problem.
In summary, the smallest natural number $n=9$.