Three, (50) Find the maximum value of $n$ such that there are $n$ points in the plane, where among any three points, there must be two points whose distance is 1.
If there exist $n(n \geqslant 8)$ points satisfying the conditions of the problem, let $V=\left\{v, v_{1}, v_{2}, \cdots, v_{7}\right\}$ represent any eight of these points. When and only when the distance between two points is 1, connect an edge between these two points, forming a graph $G$.

If there exists a point (let's assume it is $v$) with a degree less than or equal to 3, then at least four points not connected to $v$ must be pairwise connected, which is clearly impossible.
Therefore, the degree of any point is at least 4.
In the point set $V=\left\{v, v_{1}, v_{2}, \cdots, v_{7}\right\}$, take a point (let's assume it is $v$) at the vertex of the convex hull $T$. Point $v$ is connected to $v_{1}, v_{2}, v_{3}, v_{4}$ (where $v_{1}, v_{2}, v_{3}, v_{4}$ are arranged in a counterclockwise direction). Then $\left|v_{1} v_{4}\right| \neq 1$ (otherwise, $v_{2}, v_{3}, v_{4}$ would have no two points connected, which is a contradiction). Therefore, $v_{2}, v_{3}$ must each be connected to at least one of $v_{1}, v_{4}$. Thus, $v_{2}, v_{3}$ are each connected to exactly one of $v_{1}, v_{4}$ (otherwise, assume $v_{2}$ is connected to both $v_{1}, v_{4}$, then $v_{1}, v_{3}, v_{4}$ would have no two points connected, which is a contradiction).

Moreover, $v_{2}, v_{3}$ cannot both be connected to the same point among $v_{1}, v_{4}$. Therefore, among the points $v_{1}, v_{2}, v_{3}, v_{4}$, only $\left|v_{1} v_{2}\right|=\left|v_{3} v_{4}\right|=1$ or $\left|v_{1} v_{3}\right|=\left|v_{2} v_{4}\right|=1$ (with $\left.\left|v_{2} v_{3}\right| \neq 1\right)$ can occur.

Thus, each of $v_{1}, v_{2}, v_{3}, v_{4}$ must be connected to at least two of $v_{5}, v_{6}, v_{7}$. Therefore, the points $v_{5}, v_{6}, v_{7}$ must draw at least eight edges to $v_{1}, v_{2}, v_{3}, v_{4}$. By the pigeonhole principle, at least one point (let's assume it is $v_{5}$) must draw at least three edges to $v_{1}, v_{2}, v_{3}, v_{4}$, meaning there exist three points among $v_{1}, v_{2}, v_{3}, v_{4}$ that are all 1 unit away from both $v$ and $v_{5}$, which is impossible.
Therefore, $n \leqslant 7$.
Take two rhombuses $A B C D$ and $A B_{1} C_{1} D_{1}$ with side lengths of 1, such that
$$
\angle B A D=\angle B_{1} A D_{1}=60^{\circ}, C C_{1}=1 .
$$

It is easy to prove that among $A, B, C, D, B_{1}, C_{1}, D_{1}$, any three points have at least two points whose distance is 1.

Therefore, the maximum value of $n$ is 7.