Three. (50 points) Consider the 60 vertices of a regular 60-gon, where 10 vertices are colored red, 20 vertices are colored yellow, and the remaining 30 vertices are colored blue. Find the smallest positive integer $m$, such that it is always possible to find $m$ lines that do not pass through any vertices and divide the plane into several regions, each of which contains points of the same color.
Three, the minimum value of $m$ is 30.
On the one hand, provide an example to show that fewer than 30 lines cannot ensure the division as described in the problem.
Let $R$ represent red, $Y$ represent yellow, and $B$ represent blue. Then, the vertices of the regular 60-gon are colored in the sequence
RBYBYBRBYBYB $\cdots R B Y B Y B$
Since the two endpoints of each edge have different colors, at least one line must pass through each edge. Because each line can pass through at most two edges, fewer than 30 lines cannot achieve the division as described in the problem under this coloring method.
Therefore, $m \geqslant 30$.
On the other hand, show that 30 lines can definitely complete the division as required by the problem.
In fact, for any vertex, we can select two points close to it on the two edges it lies on and draw a line such that the vertex and the other 59 vertices are on opposite sides of this line. By drawing 30 lines for the 30 red and yellow points in this manner, each resulting region will contain points of only one color. Thus, these 30 lines can complete the required division.
In conclusion, the minimum value of $m$ is 30.