2. A set of 4027 points in the plane is called a "Colombian point set", where no three points are collinear, and 2013 points are red, 2014 points are blue. Draw a set of lines in the plane, which can divide the plane into several regions. If a set of lines for a Colombian point set satisfies the following two conditions, it is called a "good line set":
(1) These lines do not pass through any point in the Colombian point set;
(2) No region contains points of both colors.

Find the minimum value of $k$, such that for any Colombian point set, there exists a good line set consisting of $k$ lines.
2. $k=2013$.
Solution 1 First, give an example to show that $k \geqslant 2013$.
Mark 2013 red points and 2013 blue points alternately on a circle, and color another point in the plane blue. This circle is divided into 4026 arcs, each with endpoints of different colors. If the requirement of the problem is to be met, each arc must intersect with some drawn line. Since each line can intersect the circle at most twice, at least $\frac{4026}{2}=2013$ lines are needed.
Next, prove that 2013 lines can meet the requirement.
Notice that for any two points of the same color $A$ and $B$, two lines can be used to separate them from other points. The method is: draw two lines parallel to $AB$ on either side of $AB$. As long as they are sufficiently close to $AB$, the strip between them will contain only the two colored points $A$ and $B$.

Let $P$ be the convex hull of all colored points, and consider the following two cases.
(1) Assume $P$ has a red vertex, denoted as $A$. Then a line can be drawn to separate point $A$ from all other colored points. This way, the remaining 2012 red points can be paired into 1006 pairs, and each pair can be separated from all other colored points using two parallel lines. Therefore, a total of 2013 lines can meet the requirement.
(2) Assume all vertices of $P$ are blue. Consider two adjacent vertices on $P$, denoted as $A$ and $B$. Then a line can be drawn to separate these two points from all other colored points. This way, the remaining 2012 blue points can be paired into 1006 pairs, and each pair can be separated from all other colored points using two lines. Therefore, a total of 2013 lines can meet the requirement.

[Note] The convex hull can be ignored, and only a line passing through two colored points $A$ and $B$ can be considered, such that all other colored points are on one side of this line. If $A$ and $B$ include a red point, then proceed as in (1); if $A$ and $B$ are both blue, then proceed as in (2).

If $m$ is odd, then there exists an $N$ such that for any $m \leqslant n \leqslant N$, $f(m, n)=m$.
For any $n>N$, $f(m, n)=m+1$.
Solution 2 Another proof that 2013 lines can meet the requirement.
First, give a more general conclusion: If there are $n$ marked points in the plane with no three points collinear, and these points are arbitrarily colored red or blue, then $\left[\frac{n}{2}\right]$ lines can meet the requirement of the problem, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Use induction on $n$.
When $n \leqslant 2$, the conclusion is obvious.
Assume $n \geqslant 3$.
Consider a line passing through two colored points $A$ and $B$ such that all other colored points are on one side of this line. For example, an edge of the convex hull of all colored points is such a line.
Temporarily remove points $A$ and $B$ from consideration. By the induction hypothesis, the remaining points can be separated using $\left[\frac{n}{2}\right]-1$ lines. Now, re-include points $A$ and $B$, and consider three cases.
(1) If points $A$ and $B$ are the same color, then a line parallel to $l$ can be drawn to separate points $A$ and $B$ from other colored points. Clearly, the $\left[\frac{n}{2}\right]$ lines thus obtained can meet the requirement.
(2) If points $A$ and $B$ are different colors, but are separated by some already drawn line, then the line parallel to $l$ also meets the requirement.
(3) If points $A$ and $B$ are different colors and are in the same region after drawing the $\left[\frac{n}{2}\right]-1$ lines. By the induction hypothesis, at least one color will have no other colored points in that region. Without loss of generality, assume the only blue point in that region is $A$. Then, a line can be drawn to separate point $A$ from all other colored points.
Thus, the induction step is completed.
[Note] Generalize the problem by replacing 2013 and 2014 with any positive integers $m$ and $n$, assuming $m \leqslant n$. Denote the solution to the corresponding problem as $f(m, n)$.
Following the idea of Solution 1, we get $m \leqslant f(m, n) \leqslant m+1$.
If $m$ is even, then $f(m, n)=m$.