The answer is any integer $n \ge 6$.
When $n = 6$, consider two equilateral triangles of side lengths 1 having parallel sides and the same orientation such that each pair of corresponding vertices is at a distance 1 apart. It is clear that each point has a distance 1 to three other points, so that it is the circumcentre of some triangle.

Assume there are $n$ distinct points such that each point is the circumcentre of some triangle. Consider the convex hull of these $n$ points. Let $AB$ be a side of the convex hull. We may assume there is no point on the segment $AB$. Let $C$ be a point such that $\angle ACB$ is the largest possible. Note that $\angle ACB > 0^\circ$ since it is not possible that all $n$ points are collinear.
If $\angle ACB \ge 90^\circ$, then the circumcentre $O$ of $\triangle ABC$ lies on $AB$ or on a different side to the line $AB$ as $C$. In both cases, $O$ is distinct from all these points. We can take $O$ as the $(n+1)$st point so that each point is the circumcentre of some triangle.

If $\angle ACB < 90^\circ$, then the circumcentre $O$ of $\triangle ABC$ satisfies
$$\angle AOB = 2\angle ACB > \angle ACB.$$
From the choice of $C$, the point $O$ must be distinct from all the points. Again, we can take $O$ as the $(n+1)$st point.
By induction, we find that the property is satisfied for any $n \ge 6$.
It remains to show that $n = 5$ does not satisfy the property. If this is true, then both $n = 3$ and $n = 4$ do not satisfy the property from the above inductive step.
Let the points be $A, B, C, D, E$. WLOG assume $A$ is the circumcentre of $\triangle BCD$. One of the following must occur.
* If $B$ is the circumcentre of $\triangle ACD$, then $AC = AD = AB = BC = BD$. So both $\triangle ABC$ and $\triangle ABD$ are equilateral triangles. Note that at least one of $A, B$ must lie on a circle with centre $C$ that contains 3 points. So $E$ lies on the circle with centre $C$ that passes through $A$ and $B$. Similarly, $E$ lies on the circle with centre $D$ that passes through $A$ and $B$. This is a contradiction since the two circles only meet at $A$ and $B$.

* If $B$ is the circumcentre of $\triangle ACE$, then $AC = AD = AB = BC = BE$. So $\triangle ABC$ is an equilateral triangle. Thus, there must be 3 points lying on the circle with centre $C$ that passes through $A$ and $B$. In either case $CA = CB = CD$ or $CA = CB = CE$, we get the same configuration as above and hence it is a contradiction.
* If $B$ is the circumcentre of $\triangle CDE$, we may also assume $C$ is the circum-centre of $\triangle BDE$ since otherwise it is included in the above two cases by symmetry. Then this is exactly the same as the first case. Again, this yields a contradiction.
Therefore, $n = 5$ does not satisfy the property. The result then follows.