$5 \cdot 84$ A host is waiting for either 7 or 11 children to arrive and has prepared 77 marbles as gifts. He divides these marbles into $n$ bags so that each child, whether 7 or 11, can receive several bags of marbles, and the 77 marbles are evenly distributed among these children. Find the minimum value of $n$.


[Solution] Since these $n$ bags of marbles are divided into 7 portions, each with 11 marbles, we use 7 vertices $x_{1}, x_{2}, \cdots, x_{7}$ to represent them.

These $n$ bags of marbles can also be divided into 11 portions, each with 7 marbles, which we represent using 11 vertices $y_{1}, y_{2}, \cdots, y_{11}$.

If a certain bag of marbles appears in both $x_{i}$ and $y_{i}$, we draw an edge between $x_{i}$ and $y_{i}$. This results in a bipartite graph
$$
G=(X, Y, E),
$$

where $X=\left\{x_{1}, x_{2}, \cdots, x_{7}\right\}, Y=\left\{y_{1}, y_{2}, \cdots, y_{11}\right\}$.
This bipartite graph $G$ must be connected.
If not, let $G^{\prime}=\left(X^{\prime}, Y^{\prime}, E\right)$ be a connected subgraph, and let $\left|X^{\prime}\right|=a,\left|Y^{\prime}\right|=$ $b$, where $a \leqslant 7, b \leqslant 11$, and the equalities cannot hold simultaneously.
Since $G^{\prime}$ is connected, we have
$$
11 a=7 b .
$$

Since $(7,11)=1$, it follows that
$$
a=7, b=11,
$$

which contradicts $a \leqslant 7, b \leqslant 11$ and the fact that the equalities cannot hold simultaneously.
Since $G$ is a connected graph, its number of edges $\geqslant 7+11-1=17$. Therefore,
$$
n \geqslant 17 \text {. }
$$

In fact, when $n=17$, if the number of marbles in each bag is
$$
7,7,7,7,7,7,7,4,4,4,4,3,3,3,1,1,1 \text {. }
$$

then this satisfies the problem's requirements.
Thus, the minimum value of $n$ is 17.