1+ Given $n$ points $P_{1}, P_{2}, \cdots, P_{n}$ on a plane, where no three points are collinear. Each point $P_{i}(1 \leqslant i \leqslant n)$ is arbitrarily colored either red or blue. Let $S$ be the set of triangles whose vertex set is $\left\{P_{1}, P_{2}, \cdots, P_{n}\right\}$, and has the property: for any two line segments $P_{i} P_{j}$ and $P_{u} P_{v}$, the number of triangles in $S$ that have $P_{i} P_{j}$ as a side is the same as the number of triangles in $S$ that have $P_{u} P_{v}$ as a side. Find the smallest $n$ such that in $S$ there are always two triangles, each of whose vertices have the same color.
Let the number of triangles in $S$ with $P_{i} P_{j}$ as an edge be $k$, then $k$ is a positive integer and is independent of $i$ and $j$. Since there are $\mathrm{C}_{n}^{2}$ edges $P_{i} P_{j}$, the total number of edges produced by all triangles in $S$ is $k \mathrm{C}_{n}^{2}$. Each triangle has three edges, so $|S|=\frac{1}{3} k \mathrm{C}_{n}^{2}$ (i.e., there are $\frac{1}{3} k \mathrm{C}_{n}^{2}$ triangles in total). Let $x$ be the number of triangles in $S$ with all vertices of the same color, then the number of triangles in $S$ with vertices of different colors is $\frac{1}{3} k \mathrm{C}_{n}^{2}-x$. Each triangle with vertices of different colors produces two edges with endpoints of different colors, so the total number of edges with endpoints of different colors in $S$ is $2\left(\frac{1}{3} k \mathrm{C}_{n}^{2}-x\right)$.

On the other hand, suppose $P_{1}, \cdots, P_{n}$ have $n_{1}$ points colored red and $n_{2}$ points colored blue, with $n_{1}+n_{2}=n$. By assumption, each edge with endpoints of different colors appears $k$ times in the set of all triangles in $S$, so the total number of such edges is $k n_{1} n_{2}$. Therefore,
$$
2\left(\frac{1}{3} k \mathrm{C}_{n}^{2}-x\right)=k n_{1} n_{2}
$$

Solving for $x$, we get
$$
\begin{aligned}
x & =\frac{k}{6}\left(2 \mathrm{C}_{n}^{2}-3 n_{1} n_{2}\right)=\frac{k}{6}\left(n^{2}-n-3 n_{1} n_{2}\right) \\
& \geqslant \frac{k}{6}\left(n^{2}-n-3\left(\frac{n_{1}+n_{2}}{2}\right)^{2}\right)=\frac{k}{6} n\left(\frac{n}{4}-1\right) \\
& =\frac{k}{24} n(n-4) .
\end{aligned}
$$

Thus, when $n \geqslant 8$,
$$
x \geqslant \frac{n}{24}(n-4) \geqslant \frac{8 \times 4}{24}=\frac{4}{3}>1
$$
(because $k \geqslant 1$), i.e., $x \geqslant 2$. Therefore, $n \geqslant 8$.
When $n=7$, the conclusion is not necessarily true. For example, if points $1, 2, 4$ are colored red and points $3, 5, 6, 7$ are colored blue, the set of triangles $\{\dot{1}, \dot{2}, \dot{4}\},\{\dot{2}, 3,5\},\{3, \dot{4}, 6\},\{\dot{4}, 5,7\}$, $\{5,6, \mathbf{i}\},\{6,7,2\},\{7,1,3\}$ meets the requirement (each edge $i j$ appears in one triangle), but there are no two triangles with all vertices of the same color.