Four, (50 points) Each point in the plane is colored with one of $n$ colors, while satisfying:
(1) Each color has infinitely many points, and they are not all 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.
Given $n \geqslant 4$.
If $n=4$, take a fixed circle $\odot O$ and three points $A, B, C$ on it. Color the arc $\overparen{A B}$ (including point $A$ but not $B$), the arc $\overparen{B C}$ (including point $B$ but not $C$), and the arc $\overparen{C A}$ (including point $C$ but not $A$) with colors 1, 2, and 3, respectively. Color all other points in the plane with color 4. This satisfies the condition and there do not exist four points of different colors on the same circle.
Therefore, $n \geqslant 5$.
When $n=5$, assume that there do not exist four points of different colors on the same circle. By condition (2), there exists a line $l$ with exactly two colors of points (assume $l$ has only points of colors 1 and 2). By condition (1), there exist points $A, B, C$ with colors 3, 4, and 5, respectively, that are not collinear. Let the circle passing through $A, B, C$ be $\odot O$ (as shown in Figure 3).

If $\odot O$ intersects $l$, then there exist four points of different colors on the same circle, which is a contradiction.

If $\odot O$ is disjoint from $l$, draw a perpendicular from point $O$ to $l$ intersecting $l$ at point $D$.

Assume $D$ is colored 1, and the perpendicular intersects $\odot O$ at points $E$ and $S$, as shown in Figure 3.

Assume $E$ is colored 3. Consider a point $F$ on $l$ with color 2, and let $FS$ intersect $\odot O$ at point $G$.

Since $E G \perp G F$, points $D, E, G, F$ are concyclic. Therefore, $G$ must be colored 3.

Also, one of $B, C$ must be different from $S$ (assume it is $B$), and $SB$ intersects $l$ at point $H$.

Since $E B \perp B H$, points $B, E, D, H$ are concyclic. Therefore, $H$ must be colored 1.
Thus, $S B \cdot S H = S E \cdot S D = S G \cdot S F$.
Hence, points $B, H, F, G$ are concyclic and have different colors, which is a contradiction.

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