6. Find the smallest natural number $n$, such that in any two-coloring of $K_{n}$ there are always two monochromatic triangles that share exactly one vertex.

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When $n=8$, divide 8 points into two groups, each with 4 points, connect red edges between points in the same group, and blue edges between points in different groups. In this coloring of $K_{8}$, there are no two monochromatic triangles that share exactly one vertex.

When $n \geqslant 9$, consider a $K_{9}$ formed by any 9 points. We will prove that it must have two monochromatic triangles that share exactly one vertex. Since any 2-coloring of $K_{9}$ has at least 12 monochromatic triangles, this corresponds to $12 \times 3 = 36$ vertices, so there must be a vertex $A$ that is a vertex of at least $\frac{36}{9}=4$ triangles. For each of these 4 triangles, consider the two points other than $A$. Among these 4 pairs of points, if there are two pairs with no common point, then the corresponding two monochromatic triangles share exactly one vertex. Otherwise, assume the first pair is $\left\{P_{1}, P_{2}\right\}$, the second pair is $\left\{P_{1}, P_{3}\right\}$. Consider the third pair; if it is $\left\{P_{2}, P_{3}\right\}$, then the fourth pair will be disjoint from one of the first three pairs, thus proving the point. Assume the third pair is $\left\{P_{1}, P_{4}\right\}$, and the fourth pair is $\left\{P_{1}, P_{5}\right\}$. Thus, $A P_{1}, P_{1} P_{i}, A P_{i}(2 \leqslant i \leqslant 5)$ are all the same color, say red. If any edge among $P_{2}, P_{3}, P_{4}, P_{5}$ is red, then two monochromatic triangles that share exactly one vertex have already appeared. For example, if $P_{2} P_{3}$ is red, then $\triangle A P_{2} P_{3}$ and $\triangle A P_{1} P_{4}$ are monochromatic triangles sharing vertex $A$. If all edges among $P_{2}, P_{3}, P_{4}, P_{5}$ are blue, then $\triangle P_{2} P_{3} P_{4}$ and $\triangle A P_{2} P_{1}$ share vertex $P_{2}$ and are monochromatic triangles. In summary, $K_{\min }=9$.