Four, (50 points) The Changchun Carnival $G$ has a total of 101 attractions, and any two attractions can always be connected by some bridges, with each bridge connecting exactly two attractions, and there is at most one bridge between any two attractions. If starting from a certain attraction, one can travel along a series of different bridges through different attractions and return to the starting attraction, then such a series of bridges is called a "loop," and the number of bridges in the loop is defined as the "length" of the loop. If every bridge in the carnival $G$ lies on a loop of length at most $c(G)$, and the total number of bridges is $e(G)$, find the minimum possible value of $c(G)+e(G)$.
The minimum value sought is 121.
For example, when $G$ is a graph consisting of 10 cycles of length 11 sharing a common vertex,
$$
\begin{array}{l}
c(G)=11, e(G)=110 \\
\Rightarrow c(G)+e(G)=121 .
\end{array}
$$

Below, we show that $c(G)+e(G) \leqslant 121$.
Define the cycles $C_{1}, C_{2}, \cdots, C_{m}$ recursively: take any edge $e_{1}$, and select a smallest cycle $C_{1}$ containing $e_{1}$.

Assume $C_{1}, C_{2}, \cdots, C_{i-1}$ have already been defined, take an edge $e_{i}$ not contained in any of the cycles $C_{1}, C_{2}, \cdots, C_{i-1}$ (if no such $e_{i}$ exists, terminate), and then select a smallest cycle $C_{i}$ containing $e_{i}$. Remove the edges $e_{m}, e_{m-1}, \cdots, e_{1}$ from the graph sequentially.

Notice that this process does not disrupt the connectivity of the graph $G$, and a tree is a minimally connected graph.
$$
\begin{array}{l}
\text { Then } e(G)-m \geqslant 101-1 \\
\Rightarrow e(G) \geqslant m+100 .
\end{array}
$$

Furthermore, the cycles $C_{1}, C_{2}, \cdots, C_{m}$ cover all the edges of the graph $G$, and $\left|C_{i}\right| \leqslant c(G)$.
$$
\begin{array}{l}
\text { Thus, } m c(G) \geqslant \sum_{i=1}^{m}\left|C_{i}\right| \geqslant e(G) . \\
\text { Hence } c(G)+e(G) \geqslant \frac{m+100}{m}+m+100 \geqslant 121,
\end{array}
$$

This completes the proof.