4. A set of $n$ points $P_{1}, P_{2}, \cdots, P_{n}$ in the plane, no three of which are collinear, is denoted as $D$. A line segment is drawn between any two points, and the lengths of all these line segments are distinct. In a triangle, the side that is neither the longest nor the shortest is called the "middle side" of the triangle; if all three sides of a triangle are middle sides (not necessarily the middle sides of this triangle), then the triangle is called a "middle side triangle" in the set $D$. A line $l$ that does not pass through any point $P_{i}(i=1,2, \cdots, n)$ divides the set $D$ into two subsets $D_{1}$ and $D_{2}$. If, regardless of how these $n$ points are distributed and how $l$ is chosen, there always exists a subset $D_{k}$ $(k \in\{1,2\})$ such that $D_{k}$ contains a middle side triangle. Find the minimum value of $n$.


4. The minimum value of $n$ is 11.

When $n \geqslant 11$, regardless of how $l$ is chosen, there always exists a subset, let's assume it is $D_{1}$, such that $D_{1}$ contains at least six points.

Consider all the middle edges of the triangles in $D_{1}$ and color them red, and color the other edges blue.

By Ramsey's theorem, we know that there must exist a monochromatic triangle. Since each triangle has a middle edge, this monochromatic triangle must be red. Therefore, there exists a middle edge triangle in $D_{1}$.

Below is the proof: When $n \leqslant 10$, there exists a point set $D$ and a line $l$ such that neither $D_{1}$ nor $D_{2}$ contains a middle edge triangle.

If $n=10$, consider the two subsets $D_{1}$ and $D_{2}$ on either side of the line $l$, each containing five points, and the distribution is the same.

Assume the five points in $D_{1}$ are $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$, and they are arranged in a counterclockwise order on the circumference of a circle, with
\[
\begin{array}{l}
\overparen{P_{1} P_{2}}=\frac{\pi}{10}, \overparen{P}_{2} P_{3}=\frac{3 \pi}{5}, \overparen{P_{3} P_{4}}=\frac{3 \pi}{10}, \\
\overparen{P_{4} P_{5}}=\frac{3 \pi}{4}, \overparen{P_{5} P_{1}}=\frac{\pi}{4}.
\end{array}
\]

Then the distances between $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ are all different, and $P_{2} P_{3}, P_{3} P_{1}, P_{1} P_{5}, P_{5} P_{4}, P_{4} P_{2}$ are middle edges, but there is no middle edge triangle.

If $n<10$, for the cases where $D_{1}$ and $D_{2}$ have fewer than five points, simply remove some points from the previous example, and there will still be no middle edge triangle.
In summary, the minimum value of $n$ is 11.