4. Let there be $n(n \geqslant 4)$ points $V_{1}, V_{2}, \cdots, V_{n}$ in a plane, with no three points collinear. Some of these points are connected by line segments. Place $n$ chess pieces, labeled $1, 2, \cdots, n$, on these $n$ points, with exactly one piece on each point. Now perform the following operation on these $n$ chess pieces: each time, select several pieces and move each of them to another point that is connected to their current point by a line segment; after the operation, there is still exactly one piece on each point, and no two pieces exchange positions before and after the operation. If a way of connecting the points with line segments allows, regardless of the initial placement of the $n$ chess pieces, to always achieve, after a finite number of operations, that each piece labeled $k(k=1,2, \cdots, n)$ is at point $V_{k}$, then this way of connecting the points is called "harmonious". Find the minimum number of line segments in all harmonious ways of connecting the points.
（Fuyun Hao, problem contributor）
4. The minimum value sought is $n+1$.

First, when the number of line segments is no more than $n$, it is easy to see that one of the following two scenarios must occur:
(1) There exists a point among $V_{1}, V_{2}, \cdots, V_{n}$ that is connected by a line segment to at most one other point;
(2) For each point among $V_{1}, V_{2}, \cdots, V_{n}$, exactly two points are connected to it by line segments.
If scenario (1) occurs, let's assume this point is $V_{1}$. Then it can be proven that the chess piece starting at point $V_{1}$ can only remain at that point. If not, suppose that in a certain operation, this chess piece moves to vertex $V_{i}$. By the condition, in this operation, there must be a chess piece on another vertex $V_{j}$ that moves to $V_{1}$. This means that $V_{i}$ and $V_{j}$ are two different points, both connected to $V_{1}$, which contradicts the assumption. As long as the chess piece at point $V_{1}$ is not the 1st chess piece, it is impossible to move this chess piece to the corresponding vertex through a finite number of operations. Therefore, such a way of connecting line segments is inharmonious.
If scenario (2) occurs, it is easy to see that all the line segments form one or more closed broken lines. If there are multiple broken lines, then there clearly exist two points $V_{i}$ and $V_{j}$ such that the chess piece on $V_{i}$ cannot move to $V_{j}$ through operations. Hence, as long as the chess piece at point $V_{i}$ is the $j$-th chess piece at the start, it is impossible to move this chess piece to the corresponding vertex through a finite number of operations. Therefore, such a way of connecting line segments is inharmonious; If it is a single broken line, then let's assume that points $V_{i}$ and $V_{i+1}$ are connected by a line $(i=1,2, \cdots, n, V_{n+1}=V_{1})$. It is easy to see that in one operation, either all the chess pieces remain stationary, or the chess piece at point $V_{i}$ moves to point $V_{i+1}$, or the chess piece at point $V_{i+1}$ moves to point $V_{i} (i=1,2, \cdots, n)$. As long as the chess pieces at points $V_{1}$ and $V_{2}$ are the 1st and 3rd chess pieces at the start, it is impossible to move these chess pieces to the corresponding vertices through a finite number of operations. Therefore, such a way of connecting line segments is inharmonious.
On the other hand, if line segments are connected between points $V_{i}$ and $V_{i+1} \left(i=1,2, \cdots, n, V_{n+1}=V_{1}\right)$, and a line segment is connected between points $V_{n-1}$ and $V_{1}$, then it can be proven that such a way of connecting line segments is harmonious.
In fact, under such a connection method, the following two operations can be performed.
Operation $M_{1}$: Move the chess piece at point $V_{i}$ to point $V_{i+1}$ $(i=1,2, \cdots, n)$.
Operation $M_{2}$: Move the chess piece at point $V_{i}$ to point $V_{i+1}$ $(i=1,2, \cdots, n-2)$, move the chess piece at point $V_{n-1}$ to point $V_{1}$, and leave the chess piece at point $V_{n}$ stationary.
The following proof uses mathematical induction to show: For any $k(1 \leqslant k \leqslant n)$, it is possible to move the chess pieces numbered $1,2, \cdots, k$ to points $V_{1}, V_{2}, \cdots, V_{k}$ after a finite number of operations.

When $k=1$, suppose the 1st chess piece starts at point $V_{i}$. Then performing $n-i+1$ operations $M_{1}$ will move the 1st chess piece to point $V_{1}$.

Assume that after a certain operation, the chess pieces numbered $1,2, \cdots, k$ are at points $V_{1}, V_{2}, \cdots, V_{k}$, and suppose the $(k+1)$-th chess piece is at point $V_{r}(k+1 \leqslant r \leqslant n)$. Then perform $n-r$ operations $M_{1}$, followed by $r-k-1$ operations $M_{2}$, and finally $k+1$ operations $M_{1}$, which will move the chess pieces numbered $1, 2, \cdots, k+1$ to points $V_{1}, V_{2}, \cdots, V_{k+1}$.

By mathematical induction, it is ultimately possible to move each chess piece numbered $k$ to point $V_{k}(k=1,2, \cdots, n)$ after a finite number of operations. Therefore, such a way of connecting line segments is harmonious.

In summary, among all harmonious ways of connecting line segments, the minimum number of line segments is $n+1$.