3. For a finite graph, the following operation can be performed: choose any cycle of length 4, select any edge in this cycle, and remove it from the graph. For a fixed integer $n(n \geqslant 4)$, if the complete graph with $n$ vertices is repeatedly subjected to the above operation, find the minimum number of edges in the resulting graph $q$.


3. The minimum value is $n$.

If a graph can be obtained from a complete graph of $n$ vertices through operations, then this graph is called "permissible"; if any two points in a graph are connected by a path, then this graph is called "connected"; if the vertices of a graph can be divided into two sets $V_{1}$ and $V_{2}$, such that each edge is between a vertex in $V_{1}$ and a vertex in $V_{2}$, then this graph is called "bipartite".

Assume that operations are performed on graph $G$, which contains a cycle of length 4, $A B C D$. Deleting edge $A B$, the new graph obtained is denoted as $H$, then:
(1) If $G$ is connected, then $H$ is also connected;
(2) If $G$ is not bipartite, then $H$ is also not bipartite.
In fact, for conclusion (1), assume that there is a path $I$ connecting points $X, Y$ in graph $G$. If $I$ does not include the deleted edge $A B$, then $X$ and $Y$ are still connected by path $I$ in $H$. If $I$ includes $A B$, then in $H$, the path connecting $X$ and $Y$ can be obtained by replacing $A B$ with the three edges $A D, D C, C B$. Therefore, $H$ is also connected.

For conclusion (2), consider its contrapositive. If graph $H$ is bipartite, i.e., its vertices can be divided into two sets $V_{1}$ and $V_{2}$, such that each edge is between a vertex in $V_{1}$ and a vertex in $V_{2}$. Since $B C, C D, D A$ are edges in $H$, then $A, C$ belong to one of the sets $V_{1}$ or $V_{2}$, and $B, D$ belong to the other set. In graph $G$, if the vertices are divided into the same two sets $V_{1}$ and $V_{2}$ as in $H$, then edge $A B$ satisfies being between a vertex in $V_{1}$ and a vertex in $V_{2}$. Therefore, graph $G$ is also bipartite.

Since the complete graph $K_{n}$ is connected and $K_{n}$ is not bipartite (because it contains a triangle), any permissible graph $H$ is connected and not bipartite. In particular, $H$ contains a cycle, because a graph without a cycle is bipartite.
It is easy to see that a permissible graph has at least $n$ edges.
On the one hand, a connected graph with $n$ vertices has at least $n-1$ edges. On the other hand, a connected graph with $n$ vertices and only $n-1$ edges is a tree, and a tree does not contain a cycle. Therefore, the minimum number of edges is $n$.

Below, we prove that when $n \geqslant 4$, starting from the complete graph $K_{n}$, a graph with exactly $n$ edges can be obtained through operations. Consider a graph with $n$ vertices ($n \geqslant 4$), with vertices $A_{1}, A_{2}, \cdots, A_{n-3}, X, Y, Z$, where $Z$ is connected to each other vertex by an edge, and $X$ and $Y$ are connected by an edge. This graph can be obtained from the complete graph $K_{n}$ by repeatedly performing the following operations:

From the cycle $A_{i} X Y A_{j}$, delete edge $A_{i} A_{j}$, where $1 \leqslant i < j \leqslant n-3$; from the cycle $X Y Z A_{i}$, delete edge $X A_{i}$, where $1 \leqslant i \leqslant n-3$; from the cycle $Y X Z A_{j}$, delete edge $Y A_{j}$, where $1 \leqslant j \leqslant n-3$.
In summary, the minimum number of edges is $n$.