4. In the plane of Camelot, King Arthur built a maze $\mathscr{L}$ consisting of $n$ walls, each wall being a straight line (infinitely extended), with no two walls parallel and no three walls having a common point. The wizard Merlin painted one side of each wall entirely red and the other side entirely blue.

At the intersection of two walls, there are four corners: two opposite angles are formed by one red edge and one blue edge, one angle is formed by two red edges, and one angle is formed by two blue edges. At each such intersection, there is a two-way door leading to the two opposite angles formed by edges of different colors.

After Merlin has painted each wall, the witch Morgana arranges some knights in the maze $\mathscr{B}$, who can pass through the doors but not over the walls.

Let $k(\mathscr{D})$ be the maximum value of the positive integer $k$ such that, regardless of how Merlin colors the walls of the maze $\mathscr{L}$, Morgana can always arrange at least $k$ knights so that no two knights ever meet. For each positive integer $n$, find all possible values of $k(\mathscr{D})$, where the maze $\mathscr{E}$ has $n$ walls.
4. Regardless of the shape of the maze $\mathscr{L}$, $k$ has a unique possible value $k=n+1$.

First, use mathematical induction to prove that $n$ walls divide the plane into $\mathrm{C}_{n+1}^{2}+1$ regions.
When $n=0$, since there are no walls, the entire plane constitutes one region, so the conclusion holds.

Assume that the conclusion holds for $n-1\left(\in \mathbf{Z}_{+}\right)$ walls. When the maze $\mathscr{S}$ has $n$ walls, the $n$-th wall intersects with each of the $n-1$ walls exactly once. These $n-1$ intersection points divide the $n$-th wall into $n$ segments, and each segment divides one of the regions formed by the other $n-1$ walls into two regions, thus adding $n$ regions. By the induction hypothesis, the number of regions is
$$
\left(\mathrm{C}_{n}^{2}+1\right)+n=\mathrm{C}_{n+1}^{2}+1 \text {. }
$$

Thus, the proof by mathematical induction is completed.
Let $G$ be a graph where each vertex corresponds to one of the $\mathrm{C}_{n+1}^{2}+1$ regions. If two regions are connected by a bidirectional door, then an edge is drawn between the corresponding vertices.

Next, prove that regardless of how Merlin colors the walls of the maze $\mathscr{L}$, Morgana can always arrange $n+1$ knights to meet the required conditions.

Notice that, regardless of the coloring, there are exactly $\mathrm{C}_{n}^{2}$ intersection points, and each intersection point has exactly one door. Each intersection point corresponds to an edge in the graph $G$. Adding these edges one by one, since each added edge can reduce the number of connected components by at most one, the number of connected components in the graph $G$ is at least
$\mathrm{C}_{n+1}^{2}+1-\mathrm{C}_{n}^{2}=n+1$.
If Morgana arranges the knights in regions corresponding to different connected components in the graph $G$, then no two knights will ever meet.

Here is an example: regardless of the shape of the maze $\mathscr{L}$, Merlin can color the walls of the maze $\mathscr{L}$ such that the graph $G$ has exactly $n+1$ connected components. Thus, Morgana can arrange at most $n+1$ knights.

Establish a Cartesian coordinate system such that no wall is parallel to the coordinate axes. Merlin colors the west side of each wall red and the east side blue. For each region, label it with the number of walls that have the region on their east side, which is an integer in the set $\{0,1, \cdots, n\}$. For each $i \in\{0,1, \cdots, n\}$, the regions labeled $i$ are connected by bidirectional doors. In fact, for each $i \in\{0,1, \cdots, n\}$, there is a unique region labeled $i$ that is unbounded to the north.

Consider a knight placed in a region labeled $i$ and have him walk north (if necessary, he can walk east or west along the northern boundary of the region). The knight will not get stuck because each region is convex. If this is a northern boundary, it has a unique northernmost vertex, and there is a door leading north to another region labeled $i$. Eventually, he will reach a region that is unbounded to the north and is the unique region labeled $i$. Thus, each region labeled $i$ is connected to this special region. Therefore, all regions labeled $i$ are mutually connected. This shows that there are exactly $n+1$ connected components, and Morgana can arrange at most $n+1$ knights.