Four. (25 points) Each point on 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 4 points of different colors that are concyclic.
(Zhang Limin, problem contributor)
Obviously, $n \geqslant 4$.
If $n=4$, take a fixed circle $\odot O$ and three points $A, B, C$ on it. Color the arcs $\overparen{A B}$ (including $A$ but not $B$), $\overparen{B C}$ (including $B$ but not $C$), and $\overparen{C A}$ (including $C$ but not $A$) with colors $1, 2,$ and $3$ respectively, and color all other points in the plane with color $4$. This satisfies the conditions and there do not exist four points of different colors on the same circle.
Therefore, $n \neq 4, n \geqslant 5$.
When $n=5$, by condition (2), there exists a line $l$ on which there are exactly two colors of points. Without loss of generality, assume that the points on line $l$ are only of colors $1$ and $2$. By condition (1), there exist points $A, B, C$ of colors $3, 4, 5$ respectively that are not collinear. Let the circle passing through points $A, B, C$ be $\odot O$.

If $\odot O$ intersects line $l$, then there exist four points of different colors on the same circle;

If $\odot O$ is disjoint from line $l$ and $\odot O$ has points of colors $1$ and $2$, then there exist four points of different colors on the same circle;

If $\odot O$ is disjoint from line $l$ and $\odot O$ has no points of colors $1$ and $2$, as shown in Figure 2, 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 $3$, and consider a point $F$ on line $l$ with color $2$. The line $FS$ intersects $\odot O$ at point $G$. Since $EG \perp GF$, points $D, E, F, G$ are concyclic.

If $G$ is not a point of color $3$, then there exist four points of different colors on the same circle;

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

If $H$ is a point of color $2$, then $B, H, D, E$ are four points of different colors on the same circle;
If $H$ is a point of color $1$, since
$$
SB \cdot SH = SE \cdot SD = SG \cdot SF,
$$
points $B, H, F, G$ are concyclic.
Thus, $B, H, F, G$ are four points of different colors on the same circle.

In summary, when $n=5$, there exist four points of different colors on the same circle.
Therefore, the minimum value of $n$ is 5.