There are $n \geqslant 5$ distinct points on the plane, each point is exactly at a distance of 1 from 4 other points, find the minimum value of such $n$.

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Let the $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, and let $A_{1} A_{i}=1 (i=2,3,4,5)$.
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, at least one of $A_{2} A_{3}, A_{2} A_{4}, A_{2} A_{5}$ is not 1.
Also, $A_{2}$ is exactly 1 unit away from 4 points $A_{i}$. Therefore, $n \geqslant 6$.
The circle with center $A_{1}$ and radius 1 intersects the circle with center $A_{6}$ and radius 1 at most at two points, so at most two of $A_{6} A_{2}, A_{6} A_{3}, A_{6} A_{4}, A_{6} A_{5}$ are 1. Therefore, $A_{6}$ is at least 1 unit away from two points among $A_{7}, \cdots, A_{n}$, so $n \geqslant 8$.

When $n=8$, there must be $A_{6} A_{7}=1, A_{6} A_{8}=1$. Similarly, $A_{7} A_{8}=1$. Therefore, each of $A_{6}, A_{7}, A_{8}$ is exactly 1 unit away from two points among $A_{2}, A_{3}, A_{4}, A_{5}$.

The circles with centers $A_{6}, A_{7}, A_{8}$ and radius 1 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 $A_{3}$, which is a contradiction.
Therefore, $n \geqslant 9$. $n=9$ can be achieved, as shown in the figure on the right.