Three. (50 points) On a plane, there are $n(n \geqslant 5)$ distinct points, each point is exactly at a distance of 1 from four other points. Find the minimum value of such $n$.


Three, let the $n$ points be $A_{1}, A_{2}, \cdots, A_{n}, A_{1} A_{i}=1$ $(i=2,3,4,5)$.

Since the circle with center $A_{1}$ and radius 1 intersects the circle with center $A_{2}$ and radius 1 at most at two points, thus, among $A_{2} A_{3}, A_{2} A_{4}, A_{2} A_{5}$, at least one is not 1. Also, $A_{2}$ is exactly 1 unit away from four $A_{i}$, hence, $n \geqslant 6$.

Since the circle with center $A_{1}$ and radius 1 intersects the circle with center $A_{6}$ and radius 1 at most at two points, thus, among $A_{6} A_{2}, A_{6} A_{3}, A_{6} A_{4}, A_{6} A_{5}$, at most two are 1. Also, $A_{6}$ is 1 unit away from at least two points among $A_{7}, \cdots, A_{n}$, thus, $n \geqslant 8$.
When $n=8$, there must be $A_{6} A_{7}=1, A_{6} A_{8}=1$.
Similarly, $A_{0}=1$.
Thus, each of $A_{6}, A_{7}, A_{8}$ is exactly 1 unit away from two points among $A_{2}, A_{3}, A_{4}, A_{5}$.

Circles with radius 1 and centers at $A_{6}, A_{7}, A_{8}$ intersect at three other points, at least two of which are among $A_{2}, A_{3}, A_{4}, A_{5}$, say $A_{2}, A_{3}$.

Assume these two points are on the circle with center $A_{6}$, then $A_{2}, A_{6}, A_{3}$ are collinear, and $A_{2} A_{3}$ is a diameter. This way, $A_{1}$ cannot be 1 unit away from both $A_{2}$ and $A_{3}$, a contradiction.
Therefore, $n \geqslant 9$.
$n=9$ can be achieved, as shown in Figure 5.