4. Find the integer $n(n \geqslant 5)$: There are exactly $n$ distinct points in the plane, such that each point is the circumcenter of a triangle formed by three of the $n$ points.
4. $n$ is any integer not less than 6.

When $n=6$, consider the two equilateral triangles shown in Figure 2: both have a side length of 1, corresponding sides are parallel and in the same direction, and the distance between each pair of corresponding vertices is 1.

Clearly, each point is 1 unit away from the other three points. Thus, each point is the circumcenter of some triangle.
Assume there are $n (n \geqslant 6)$ distinct points that satisfy the condition.
Consider the convex hull of these $n$ points.
As shown in Figure 3, let $A B$ be one of the edges of this convex hull. We can assume that there are no other points of these $n$ points on the segment $A B$.
If $\angle A C B \geqslant 90^{\circ}$.
By the selection of point $C$, point $O$ must not coincide with any of these points. Take $O$ as the $(n+1)$-th point.
By induction, we know that $n \geqslant 6$ satisfies the condition.
Now, we show that $n=5$ does not satisfy the condition.
If $n=5$ satisfies the condition, then by the previous inductive discussion, we know that $n=3,4$ do not satisfy the condition.
Let these five points be $A, B, C, D, E$.
Without loss of generality, assume $A$ is the circumcenter of $\triangle B C D$.
We consider three cases.
(i) If $B$ is the circumcenter of $\triangle A C D$, then $A C = A D = A B = B C = B D$.
As shown in Figure 4.
Thus, $\triangle A B C$ and $\triangle A B D$ are both equilateral triangles.
Note that at least one of the points $A, B$ lies on the circle $\odot C$ with radius $C A$ (which already has three other points on it).
Therefore, point $E$ lies on the circle $\odot C$ passing through $A, B$.
Similarly, point $E$ lies on the circle $\odot D$ passing through $A, B$.
This contradicts the fact that $\odot C$ and $\odot D$ intersect at only two points $A, B$.
(ii) If $B$ is the circumcenter of $\triangle A C E$, then
$A C = A D = A B = B C = B E$.
Thus, $\triangle A B C$ is an equilateral triangle.
Therefore, there must be three points on the circle $\odot C$ passing through $A, B$.

In the cases where $C A = C B = C D$ or $C A = C B = C E$, we can similarly obtain a contradiction as in (i).
(iii) If $B$ is the circumcenter of $\triangle C D E$, assume $C$ is the circumcenter of $\triangle B D E$ (otherwise, it would become the cases (i) or (ii)), then this is exactly the case (i). Contradiction.
Therefore, $n=5$ does not satisfy the condition.
Let $C$ be the point, excluding $A$ and $B$, such that $\angle A C B$ is the largest.
By the fact that the $n$ points are not collinear, we know that $\angle A C B > 0^{\circ}$.
If $\angle A C B \geqslant 90^{\circ}$, then the circumcenter $O$ of $\triangle A B C$ is on $A B$ or on the opposite side of $C$ with respect to $A B$.

In both cases, $O$ does not coincide with any of these points. Take $O$ as the $(n+1)$-th point, such that each point is the circumcenter of some triangle.