The 21st question, given a positive integer $\mathrm{n} \geq 3$, if the $\mathrm{n}$-order simple graph $\mathrm{G}(\mathrm{V}, \mathrm{E})$ is connected, and removing all edges on any cycle in $\mathrm{G}$ makes $\mathrm{G}$ no longer connected, try to find the maximum possible value of $|\mathrm{E}(\mathrm{G})|$.
The 21st question,
Solution: Let $V(G)=\left\{v_{1}, v_{2}, \ldots, v_{n}\right\}$.
On one hand, take
$$
E(G)=\left\{v_{1} v_{2}\right\} \cup\left\{v_{1} v_{i} \mid 3 \leq i \leq n\right\} \cup\left\{v_{2} v_{i} \mid 3 \leq i \leq n\right\}
$$

This graph $G$ satisfies the given conditions, and $|\mathrm{E}(\mathrm{G})|=2 \mathrm{n}-3$.
On the other hand, we prove: For an $n$-order simple connected graph $\mathrm{G}(\mathrm{V}, \mathrm{E})$, if $|\mathrm{E}| \geq 2 \mathrm{n}-2$, it is always possible to remove all edges of a cycle in $\mathrm{G}$ such that the remaining graph is still connected.
We use induction on $\mathrm{n}$.
When $n=3, 4$, the conclusion is obviously true.
Assume the conclusion holds for $\mathrm{n}$. Now consider the case for $\mathrm{n}+1$.
Without loss of generality, assume $|\mathrm{E}|=2(\mathrm{n}+1)-2=2 \mathrm{n}$. Otherwise, if $|\mathrm{E}|>2 \mathrm{n}$, we can remove one edge from a cycle each time until $|E|=2 n$.
By the pigeonhole principle, there must exist a vertex with degree no more than $\left[\frac{4 \mathrm{n}}{\mathrm{n}+1}\right]=3$. Let this vertex be $\mathrm{v}_{1}$. We consider three cases: (i) If $\mathrm{d}\left(\mathrm{v}_{1}\right)=1$, then $\mathrm{G}-\mathrm{v}_{1}$ is a connected graph with $\mathrm{n}$ vertices and $2 \mathrm{n}-1$ edges. According to the induction hypothesis, we can remove all edges of a cycle in $\mathrm{G}-\mathrm{v}_{1}$ such that the remaining graph is still connected, and thus the original graph is also connected. The conclusion holds.
(ii) If $d\left(v_{1}\right)=2$, let $N\left(v_{1}\right)=\left\{v_{2}, v_{3}\right\}$.
(1) If $v_{2}$ and $v_{3}$ are connected, then $G-v_{1}$ is a connected graph with $n$ vertices and $2 n-2$ edges. According to the induction hypothesis, we can remove all edges of a cycle in $\mathrm{G}-\mathrm{v}_{1}$ such that the remaining graph is still connected, and thus the original graph is also connected. The conclusion holds.
(2) If $v_{2}$ and $v_{3}$ are not connected, we consider two sub-cases:
(1) Removing vertex $v_{1}$ and the two edges associated with $v_{1}$ leaves a connected graph. Let the remaining graph be $G^{\prime}$. Then $G^{\prime}$ is a connected graph with $n$ vertices and $2 \mathrm{n}-2$ edges. According to the induction hypothesis, we can remove all edges of a cycle in $G^{\prime}$ such that the remaining graph is still connected, and thus the original graph is also connected. The conclusion holds.
(2) Removing vertex $v_{1}$ and the two edges associated with $v_{1}$ leaves a disconnected graph. We add an edge between $v_{2}$ and $v_{3}$, and let the resulting graph be $\mathrm{G}^{\prime}$. Clearly, $\mathrm{G}^{\prime}$ is a connected graph with $\mathrm{n}$ vertices and $2 \mathrm{n}-1$ edges. According to the induction hypothesis, we can remove all edges of a cycle in $\mathrm{G}^{\prime}$ (clearly, edge $\mathrm{v}_{2} \mathrm{v}_{3}$ is a bridge and does not belong to any cycle), such that the remaining graph is still connected, and thus the original graph is also connected (i.e., using the path $v_{2} v_{1} v_{3}$ in the original graph to replace the edge $v_{2} v_{3}$ in $G^{\prime}$). The conclusion holds.
(iii) If $d\left(v_{1}\right)=3$, let $N\left(v_{1}\right)=\left\{v_{2}, v_{3}, v_{4}\right\}$.
If $v_{2}, v_{3}, v_{4}$ are connected by 3 edges, then removing the cycle formed by these three edges leaves a connected graph. The conclusion holds.
If $v_{2}, v_{3}, v_{4}$ are connected by at most 2 edges. Now we add edges to make $v_{2}, v_{3}, v_{4}$ all connected (at least one edge is added), and then remove $v_{1}$ and the edges associated with $v_{1}$. Let the resulting graph be $G^{\prime}$. Then $G^{\prime}$ is a connected graph with $n$ vertices and at least $2 n-2$ edges. According to the induction hypothesis, we can remove all edges of a cycle in $G^{\prime}$ such that the remaining graph is still connected, and thus the original graph is also connected (i.e., using the paths $v_{3} v_{1} v_{4}$, $v_{2} v_{1} v_{3}$, and $v_{2} v_{1} v_{4}$ in the original graph to replace the edges $v_{3} v_{4}$, $v_{2} v_{3}$, and $v_{2} v_{4}$ in $G^{\prime}$, respectively). The conclusion holds.
In summary, the maximum value of $|E(G)|$ is $2 n-3$.