The 23rd question: 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 positive number $\mathrm{n}(\mathrm{n} \geq 4)$, if the complete graph with $\mathrm{n}$ vertices is repeatedly subjected to the above operation, find the minimum possible number of edges in the resulting graph.

For a fixed positive number $\mathrm{n}(\mathrm{n} \geq 4)$, if the complete graph with $\mathrm{n}$ vertices is repeatedly subjected to the above operation, find the minimum possible number of edges in the resulting graph.
The 23rd question,
Answer: If a graph can be obtained by operating on a complete graph with $\mathrm{n}$ vertices, then this graph is called "permissible".
Assume that the operation is performed on graph $\mathrm{G}$, which contains a cycle $\mathrm{ABCD}$ of length 4. Deleting edge $\mathrm{AB}$, the new graph obtained is denoted as $H$. Clearly, if $\mathrm{G}$ is a connected graph, then $H$ is also a connected graph.

Assume $\mathrm{H}=\mathrm{X}+\mathrm{Y}$ is a bipartite graph. Since $\mathrm{BC} 、 \mathrm{CD} 、 \mathrm{DA}$ are all edges in $\mathrm{H}$, $\mathrm{A} 、 \mathrm{C}$ must belong to the same set among $\mathrm{X} 、 \mathrm{Y}$, and $\mathrm{BD}$ belongs to the other set. At this point, adding edge $\mathrm{AB}$ back to graph $\mathrm{H}$, we get the original graph $\mathrm{G}$, which is clearly still a bipartite graph. This indicates: if $\mathrm{G}$ is not a bipartite graph, then $\mathrm{H}$ is also not a bipartite graph.

Since the initial graph $\mathrm{G}$ is a complete graph of $\mathrm{n}$ vertices, it is connected and not a bipartite graph (because it contains an odd cycle). According to the above analysis, any permissible graph $\mathrm{H}$ is connected and not a bipartite graph. If a graph is a connected graph of $\mathrm{n}$ vertices and is not a bipartite graph, it must have at least $\mathrm{n}$ edges (since it is connected, it must have at least $\mathrm{n}-1$ edges; if it has exactly $\mathrm{n}-1$ edges, it is a tree, which is a bipartite graph). Therefore, any permissible graph must have at least $\mathrm{n}$ edges.

On the other hand, we prove: when $n \geq 4$, starting from the complete graph $K_{n}$, it is possible to obtain a graph with exactly $\mathrm{n}$ edges through the operations described in the problem.

In fact, let the vertices of $K_{n}$ be $A_{1} 、 A_{2}, \ldots 、 A_{n-3} 、 X 、 Y 、 Z$. We perform the following operations: from the cycle $\mathrm{A}_{\mathrm{i}} \mathrm{XYA} \mathrm{A}_{\mathrm{j}} \mathrm{A}_{\mathrm{i}}$ delete edge $\mathrm{A}_{\mathrm{i}} \mathrm{A}_{\mathrm{j}}(1 \leq \mathrm{i}<\mathrm{j} \leq \mathrm{n}-3)$; from the cycle $\mathrm{XYZA}_{\mathrm{i}} \mathrm{X}$ delete edge $\mathrm{XA}_{\mathrm{i}}(1 \leq \mathrm{i} \leq \mathrm{n}-3)$; from the cycle $\mathrm{XYZA}_{\mathrm{j}} \mathrm{Y}$ delete edge $\mathrm{YA}_{\mathrm{j}}(1 \leq \mathrm{j} \leq \mathrm{n}-3)$. The final graph obtained has exactly $\mathrm{n}$ edges.
In summary, the minimum number of edges is $\mathrm{n}$.