10. A hostess is waiting for 7 or 11 children to arrive, and she has prepared 77 marbles as gifts. She puts 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 $n$.
10. Since these $n$ bags of sand can be divided into 7 portions, each containing 11 marbles, represented by 7 vertices $x_{1}, x_{2}, \cdots, x_{7}$. These $n$ bags of marbles can also be divided into 11 portions, each containing 7 marbles, represented by vertices $y_{1}, y_{2}, \cdots, y_{11}$. If a certain bag of marbles appears in both $x_{i}$ and $y_{i}$, then an edge is drawn between $x_{1}$ and $y$. 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 $G^{\prime}=\left(X^{\prime}, Y^{\prime}, E^{\prime}\right)$ be a connected subgraph, with $\left|X^{\prime}\right|=a,\left|Y^{\prime}\right|=b$, where $a \leqslant 7, b \leqslant 11$, and the equalities cannot hold simultaneously. Since $G^{\prime}$ is connected, we have $11a = 7b$.

Since 7 and 11 are coprime, the above equation holds only when $a=7, b=11$. This is a contradiction.

Since $G$ is a connected graph, its number of edges is at least $7 + 11 - 1 = 17$, so $n \geqslant 17$.
When $n=17$, if the number of marbles in each bag is
$7,7,7,7,7,7,7,4,4,4,4$,
$3,3,3,1,1,1$,

then the requirements are met. Therefore, the minimum value of $n$ is 17.