7. Find the smallest natural number $n$, such that in any two-coloring of $K_{n}$ there exist 5 edge-disjoint monochromatic triangles.
Solution: $n_{\min }=11$.
(1) When $n \leqslant 10$. We can take a part or all of the following graph, where there do not exist 5 monochromatic triangles without common edges (solid lines represent red, no lines represent blue).
(2) When $n=11$, establish the following three lemmas:
Lemma 1: In a two-coloring of $K_{6}$, there must be two monochromatic triangles.

Lemma 2: In a two-coloring of $K_{7}$, there must be two monochromatic triangles without common edges.

Lemma 3: In a two-coloring of $K_{11}$, there must be two monochromatic triangles with exactly one common vertex.

Lemma 1 is a well-known structure. For the proof of Lemma 2, by Lemma 1, there are two monochromatic triangles. If they have no common edges, the lemma is proved. If they do, assume without loss of generality that $\triangle A_{1} A_{2} A_{3}$ and $\triangle A_{1} A_{2} A_{4}$ are both red triangles. Among them, there must be one different from $\triangle A_{2} A_{3} A_{4}$, which must be without common edges with $\triangle A_{1} A_{2} A_{3}$ or $\triangle A_{1} A_{2} A_{4}$ (drawing a diagram, it is somewhat complex).

Proof of Lemma 3: By Lemma 2, assume there exist monochromatic triangles $\triangle A_{1} A_{2} A_{3}$, $\triangle A_{4} A_{5} A_{6}$, $\triangle A_{7} A_{8} A_{9}$ (using proof by contradiction). Take 7 points $A_{1}, A_{2}, A_{3}, A_{4}, A_{7}, A_{10}, A_{11}$, among which there will be two monochromatic triangles without common edges and no common vertices. Among them, there must be one with a common vertex with $\triangle A_{1} A_{2} A_{3}$, $\triangle A_{4} A_{5} A_{6}$, or $\triangle A_{7} A_{8} A_{9}$, leading to a contradiction!

Next, we prove that $n=11$ satisfies the condition, using proof by contradiction, assume it does not hold.

First, by Lemma 3, we can assume $\triangle A_{1} A_{2} A_{3}$ and $\triangle A_{1} A_{4} A_{5}$ are monochromatic triangles.

Consider the 6 points $A_{6}, A_{7}, A_{8}, A_{9}, A_{10}, A_{11}$. If the edges between them are all the same color, then $\triangle A_{6} A_{7} A_{8}$, $\triangle A_{6} A_{10} A_{11}$, $\triangle A_{8} A_{9} A_{10}$, $\triangle A_{1} A_{2} A_{3}$, and $\triangle A_{1} A_{4} A_{5}$ are 5 monochromatic triangles without common edges, leading to a contradiction!

If there is a monochromatic triangle $T$ that has no common edges with $\triangle A_{1} A_{2} A_{3}$ and $\triangle A_{1} A_{4} A_{5}$, and shares a vertex with at least one of them, assume without loss of generality that $A_{2} \notin T$, $A_{4} \notin T$, $A_{6} \in T$. Consider the 7 points $A_{2}, A_{4}, A_{6}, A_{8}, A_{9}, A_{10}, A_{11}$, among which there are two monochromatic triangles without common edges (assume without loss of generality that $A_{8}, A_{9}, A_{10}, A_{11} \notin T$), leading to a contradiction!

By (1), assume the edges between $A_{6}, A_{7}, A_{8}, A_{9}$ are not all the same color. Among $A_{2}, A_{4}, A_{6}, A_{7}, A_{8}, A_{9}$, there are two monochromatic triangles, and these two triangles do not contain $A_{2}, A_{4}$. Assume without loss of generality that they are $\triangle A_{6} A_{8} A_{9}$ and $\triangle A_{7} A_{8} A_{9}$, and are red, then $A_{6} A_{7}$ is blue. Consider the 6 points $A_{2}, A_{4}, A_{6}, A_{7}, A_{k}, A_{i} (k=8,9; i=10,11)$. Similarly, we get: $A_{i} A_{k}, A_{i} A_{6}, A_{i} A_{7}$ are red.

Similarly, we get $A_{10} A_{11}$ is red. This way, between $A_{6}, A_{7}, A_{8}, A_{8}, A_{10}, A_{11}$, only $A_{6} A_{7}$ is blue. Consider $\triangle A_{1} A_{2} A_{3}$, $\triangle A_{1} A_{4} A_{5}$, $\triangle A_{7} A_{10} A_{11}$, $\triangle A_{8} A_{6} A_{11}$, $\triangle A_{8} A_{9} A_{10}$, a total of 5 monochromatic triangles, leading to a contradiction!
Thus, the assumption is false, $n=11$ satisfies the condition.
In conclusion, $n_{\min }=11$.