Example 8 Each point in the plane is colored with one of $n$ colors, and the following conditions are satisfied:
(1) There are infinitely many points of each color, and they do not all lie on the same line;
(2) There is at least one line on which all points are exactly two colors.

Find the minimum value of $n$ such that there exist four points of different colors that are concyclic.
(3rd Northern Mathematical Olympiad Invitational)
Explanation: If $n=4$, then a colored plane can be constructed such that: there is exactly one point on a circle with three colors, and the rest of the points are of another color, satisfying the problem's conditions and ensuring no four points of different colors are concyclic. Therefore, $n \geqslant 5$.

When $n=5$, by condition (2), there exists a line $l$ on which exactly two colors of points lie. Without loss of generality, assume that line $l$ has only points of colors 1 and 2. By condition (1), there exist points $A$, $B$, and $C$ of colors 3, 4, and 5, respectively, that are not collinear. Let the circle passing through points $A$, $B$, and $C$ be $\odot O$.
(i) If $\odot O$ intersects line $l$, then there exist four points of different colors that are concyclic.
(ii) If $\odot O$ is separated from line $l$ and $\odot O$ has points of colors 1 and 2, then there exist four points of different colors that are concyclic.
(iii) If $\odot O$ is separated from line $l$ and $\odot O$ has no points of colors 1 and 2, as shown in Figure 11, draw a perpendicular from $O$ to line $l$ intersecting $l$ at point $D$. Assume the color of $D$ is 1, and the perpendicular intersects $\odot O$ at points $E$ and $S$. Assume the color of $E$ is 3. Consider a point $F$ on line $l$ with color 2, and let $FS$ intersect $\odot O$ at point $G$. Since $EG \perp GF$, points $D$, $E$, $G$, and $F$ are concyclic.

If $G$ is not a point of color 3, then there exist four points of different colors that are concyclic;

If $G$ is a point of color 3, and $B$, $C$ must have one point different from $S$ (assume it is $B$), $SB$ intersects line $l$ at point $H$. Since $EB \perp BH$, points $B$, $E$, $D$, and $H$ are concyclic.

If $H$ is a point of color 2, then $B$, $H$, $D$, and $E$ are four points of different colors and are concyclic;
If $H$ is a point of color 1, since
$$
SB \cdot SH = SE \cdot SD = SG \cdot SF,
$$
then $B$, $H$, $F$, and $G$ are four points of different colors and are concyclic.
In summary, when $n=5$, there exist four points of different colors that are concyclic.
Therefore, the minimum value of $n$ is 5.