4.40 Find the smallest positive integer $n$, such that in any two-coloring of $K_{n}$, there exist two monochromatic triangles of the same color with no common edge.
［Solution 1］In the $K_{7}$ shown in the right figure, the line segments that are not drawn are all red, and the drawn ones are blue line segments. It is easy to see that any two blue triangles or any two red triangles have 1 common edge. This indicates that the smallest positive integer $n \geqslant 8$.

Below, we use proof by contradiction to show that in a two-colored $K_{8}$, there must be two triangles that meet the requirements of the problem.
Assume there is a two-colored $K_{8}$ in which there are no two triangles that meet the requirements of the problem.
Since a two-colored $K_{8}$ must have a monochromatic triangle, let $\triangle A_{1} A_{2} A_{3}$ be a blue triangle. Consider the complete subgraph $K_{6}$ with vertices $\left\{A_{3}, A_{4}, A_{5}, A_{6}, A_{7}, A_{8}\right\}$. By the assumption of contradiction, it cannot have a blue triangle, so it must have a red triangle. Let's assume this red triangle shares 1 vertex with $\triangle A_{1} A_{2} A_{3}$. If not, then among the 9 lines connecting the vertices of the two triangles, there must be 5 lines of the same color. Suppose there are 5 red lines. Thus, at least one of $A_{1}, A_{2}, A_{3}$ must draw two red lines to the red triangle, thereby forming a red triangle that shares 1 vertex with the blue $\triangle A_{1} A_{2} A_{3}$. Let the red triangle be $\triangle A_{3} A_{4} A_{5}$.

Consider the two-colored $K_{5}$ with vertices $\left\{A_{2}, A_{4}, A_{6}, A_{7}, A_{8}\right\}$. By the assumption of contradiction, it does not have a monochromatic triangle. Therefore, it can be decomposed into one red and one blue circle, each with 5 edges (as shown in the figure).

Consider the two-colored $K_{6}$ with vertices $\left\{A_{1}, A_{2}, A_{4}, A_{6}, A_{7}, A_{8}\right\}$. By the assumption of contradiction, it cannot have a red triangle, so it must have two blue triangles, both sharing an edge with the blue $\triangle A_{1} A_{2} A_{3}$. Of course, this edge can only be $A_{1} A_{2}$. From this, we know that $A_{1} A_{4}$ and $A_{1} A_{6}$ are blue edges.
Now consider the $K_{6}$ with vertices $\left\{A_{2}, A_{4}, A_{5}, A_{6}, A_{7}, A_{8}\right\}$.

Similarly, we can conclude that $A_{5} A_{6}$ and $A_{5} A_{7}$ are red edges. Finally, consider the edge $A_{3} A_{6}$. If it is a blue edge, then $\triangle A_{2} A_{3} A_{6}$ and $\triangle A_{2} A_{1} A_{4}$ are two blue triangles that meet the requirements, leading to a contradiction; if $A_{3} A_{6}$ is a red edge, then $\triangle A_{3} A_{4} A_{6}$ and $\triangle A_{4} A_{5} A_{7}$ are two red triangles that meet the requirements, leading to a contradiction. Thus, it is proven that in a two-colored $K_{8}$, there must be two triangles that meet the requirements of the problem. $\square$
In summary, the smallest positive integer $n=8$.
[Solution 2] From the example at the beginning of Solution 1, we only need to prove that in a two-colored $K_{8}$, there must be two triangles that meet the requirements of the problem. For this, we first prove the following lemma.
Lemma In a two-colored $K_{8}$, there are at least 8 different monochromatic triangles.
Proof of the lemma Since $K_{8}$ has a total of $C_{8}^{3}=56$ triangles, we only need to prove that there are at most 48 bichromatic triangles (triangles with 3 edges of different colors).

We call an angle with two edges of different colors a bichromatic angle. Clearly, each bichromatic triangle has exactly two bichromatic angles.

Let the number of red edges drawn from point $A_{i}$ be $r_{i}$, so the number of blue edges drawn is $7-r_{i}, i=1$, $2, \cdots, 8$. Therefore, the number of bichromatic angles with $A_{i}$ as a vertex is $r_{i}\left(7-r_{i}\right)$. Thus, the total number of bichromatic angles in $K_{8}$ is $\sum_{i=1}^{8} r_{i}\left(7-r_{i}\right)$. Therefore, the total number of bichromatic triangles does not exceed
$$
\frac{1}{2} \sum_{i=1}^{8} r_{i}\left(7-r_{i}\right) \leqslant \frac{1}{2} \sum_{i=1}^{8} 3 \times 4=48,
$$

with equality if and only if each point draws 3 or 4 red edges and 3 or 4 blue edges. 
Returning to the proof of the problem. By the lemma, in a two-colored $K_{8}$, there are at least 8 monochromatic triangles.
(1) Suppose $K_{8}$ has 5 monochromatic triangles of the same color. Without loss of generality, assume they are all blue. If these 5 blue triangles all share an edge, then the 5 blue triangles must have a common edge. Let these 5 triangles be $\triangle A_{1} A_{2} A_{3}, \triangle A_{1} A_{2} A_{4}, \triangle A_{1} A_{2} A_{5}, \triangle A_{1} A_{2} A_{6}$, and $\triangle A_{1} A_{2} A_{7}$. Consider the $K_{5}$ with vertices $\left\{A_{3}, A_{4}, A_{5}, A_{6}, A_{7}\right\}$. If it has 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 two blue triangles with no common edge, leading to a contradiction; if it has all red edges, then there must be two red triangles with no common edge, leading to a contradiction. This shows that if $K_{8}$ has 5 monochromatic triangles of the same color, there must be two with no common edge.
(2) Suppose $K_{8}$ has exactly 4 red and 4 blue 8 monochromatic triangles. If the 4 blue triangles share an edge, say $A_{1} A_{2}$, then vertex $A_{1}$ draws 5 blue edges. By the last statement in the proof of the lemma, this means that $K_{8}$ cannot have only 8 monochromatic triangles, leading to a contradiction. Therefore, if the 4 blue triangles all share an edge, they form a blue $K_{4}$. The same applies to the 4 red triangles.

If the conclusion does not hold, when $K_{8}$ has exactly 4 red and 4 blue 8 monochromatic triangles, there must be a red $K_{4}$ and a blue $K_{4}$. Clearly, they can have at most 1 common vertex. It is easy to see that, regardless of whether they have a common vertex or not, we can always take 1 triangle from the red $K_{4}$ and 1 triangle from the blue $K_{4}$, and these two triangles will have no common vertex. Then, as in the beginning of Solution 1, we can get a new monochromatic triangle, which contradicts the assumption that there are exactly 8 monochromatic triangles. This proves that in a two-colored $K_{8}$, there must be two triangles that meet the requirements of the problem.
In summary, the smallest positive integer $n=8$.