$4 \cdot 44$ Find the smallest natural number $n$, such that in a simple graph with 10 vertices, $n$ edges, and two-colored, there always exists a monochromatic triangle or a monochromatic quadrilateral.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
[Solution] The graph on the right has 10 vertices and 30 edges, with 15 red edges and 15 blue edges. It is easy to see that there are neither monochromatic triangles nor monochromatic quadrilaterals in the graph. Therefore, the smallest natural number $n \geqslant 31$.

Below, we prove that when $n=31$, there must be a monochromatic triangle or a monochromatic quadrilateral in the graph. At this point, by the pigeonhole principle, there must be 16 edges of the same color in the graph. Therefore, it is only necessary to prove that in a simple graph with 10 vertices and 16 edges, there must be a triangle or a quadrilateral.

Since there are 32 endpoints in the 16 edges, by the pigeonhole principle, there must be a vertex that is connected to at most 3 edges. Without loss of generality, assume vertex $A_{10}$ is connected to 3 edges. Thus, there are 13 edges among the remaining 9 points. The 13 edges have 26 endpoints, and by the pigeonhole principle, there must be a vertex that is connected to at most 2 edges. Without loss of generality, assume $A_{9}$ is connected to 2 edges. Thus, the subgraph with vertices $\left\{A_{1}, A_{2}, \cdots, A_{8}\right\}$ has 11 edges. The 11 edges have 22 endpoints, and by the pigeonhole principle, there must be a vertex that is connected to at most 2 edges. Without loss of generality, assume $A_{8}$ is connected to 2 edges. Thus, the subgraph with vertices $\left\{A_{1}, A_{2}, \cdots, A_{7}\right\}$ has 9 edges. Similarly, we can assume the subgraph with vertices $\left\{A_{1}, A_{2}, \cdots, A_{6}\right\}$ has 7 edges.

Consider the subgraph with vertices $\left\{A_{1}, A_{2}, \cdots, A_{6}\right\}$, which has 7 edges. At this point, there must be a cycle in the graph. If the cycle with the most edges has 6 edges, then the 7th edge, regardless of which two vertices it connects, will always form a triangle or a quadrilateral. If the cycle with the most edges has 5 edges, then the other two edges either both come from vertices outside the cycle, or at least one is a diagonal of the pentagon. In either case, a triangle or a quadrilateral is formed. If the cycle with the most edges has at most 4 edges, the conclusion is obviously true. In summary, the smallest natural number $n=31$.