6. Find the smallest natural number $n$, such that in any simple graph with 10 vertices, $n$ edges, and two-colored edges, there always exists a monochromatic triangle or a monochromatic quadrilateral.
6. Solution: $n_{\min }=31$.

On the one hand, when $n=30$, a graph can be constructed (please complete it yourself) with 30 edges and two-colored, but without any monochromatic triangles or monochromatic quadrilaterals.

On the other hand, when $n \geqslant 31$, there must be 16 edges colored the same color $\alpha$. Consider the number of "same $\alpha$-colored angles" $\angle X Y Z$ (where $X Y$ and $Y Z$ are both $\alpha$-colored) denoted as $s$,
then $s \geqslant 8 \mathrm{C}_{3}^{e}+2 \mathrm{C}_{4}^{2}=36$, and $36+16=52>\mathrm{C}_{10}^{2}$.
Thus, one of the following (1) or (2) must hold.
(1) $\angle X Y Z$ is a "same $\alpha$-colored angle", and $X Z$ is colored $\alpha$.
(2) $\angle X Y Z$ is a "same $\alpha$-colored angle", and $\angle X U Z$ is also a "same $\alpha$-colored angle".

For (1), there exists a monochromatic triangle; for (2), there exists a monochromatic quadrilateral.
Therefore, $n_{\text {min }}=31$.