(11) (25 points) The sum of $n$ positive integers $x_{1}, x_{2}, \cdots, x_{n}$ is 2009. If these $n$ numbers can be divided into 41 groups with equal sums and also into 49 groups with equal sums, find the minimum value of $n$.


11 Let the 41 groups be $A_{1}, A_{2}, \cdots, A_{41}$, where the sum of the numbers in each group is 49, and we call such groups $A$-type groups; and the 49 groups be $B_{1}, B_{2}, \cdots, B_{49}$, where the sum of the numbers in each group is 41, and we call such groups $B$-type groups.

Clearly, each term $x_{k}$ belongs to exactly one $A$-type group and one $B$-type group, i.e., there are no common terms between groups of the same type. If two groups $A_{i}$ and $B_{j}$ have two common terms $x_{r}$ and $x_{t}$, then these two numbers can be combined into one term $x_{r} + x_{t}$, which would reduce the value of $n$. Therefore, we can assume that each pair of $A_{i}$ and $B_{j}$ has at most one common term.

Now, use points $u_{1}, u_{2}, \cdots, u_{41}$ to represent $A_{1}, A_{2}, \cdots, A_{41}$, and points $v_{1}, v_{2}, \cdots, v_{49}$ to represent $B_{1}, B_{2}, \cdots, B_{49}$. If the groups $A_{i}$ and $B_{j}$ have a common term, then connect a line between the corresponding points $u_{i}$ and $v_{j}$. Thus, we obtain a bipartite graph $G$ with exactly $n$ edges and 90 vertices. We will prove that $G$ is a connected graph.

If the largest connected component of graph $G$ is $G^{\prime}$, and the number of vertices in $G^{\prime}$ is less than 90, suppose in the component $G^{\prime}$ there are $a$ $A$-type vertices $u_{k_{1}}, u_{k_{2}}, \cdots, u_{k_{a}}$ and $b$ $B$-type vertices $v_{s_{1}}, v_{s_{2}}, \cdots, v_{s_{b}}$, where $a + b < 90$, then in the corresponding $A$-type groups $A_{k_{1}}, A_{k_{2}}, \cdots, A_{k_{a}}$ and $B$-type groups $B_{s_{1}}, B_{s_{2}}, \cdots, B_{s_{b}}$, each number $x_{i}$ in an $A$-type group $A_{k_{i}}$ must appear in some $B$-type group $B_{s_{j}}$; and each number $x_{j}$ in a $B$-type group $B_{s_{i}}$ must appear in some $A$-type group $A_{r_{j}}$ (otherwise, there would be an edge connecting to a vertex outside the component, leading to a contradiction). Therefore, the sum of the numbers in the $a$ $A$-type groups $A_{k_{1}}, A_{k_{2}}, \cdots, A_{k_{a}}$ should equal the sum of the numbers in the $b$ $B$-type groups $B_{s_{1}}, B_{s_{2}}, \cdots, B_{s_{b}}$, i.e., $49a = 41b$. This implies $41 \mid a$ and $49 \mid b$, so $a + b \geq 41 + 49 = 90$, which is a contradiction! Therefore, $G$ is a connected graph. Thus, the graph $G$ has at least $90 - 1 = 89$ edges, i.e., $n \geq 89$;

On the other hand, we can actually construct a sequence $x_{1}, x_{2}, \cdots, x_{89}$ with 89 terms that satisfies the conditions of the problem. For example, take $x_{1} = \cdots = x_{41} = 41, x_{42} = \cdots = x_{75} = 8, x_{76} = \cdots = x_{79} = 7, x_{80} = \cdots = x_{83} = 1, x_{84} = x_{85} = 6, x_{86} = x_{87} = 2, x_{88} = 5, x_{89} = 3$ (this sequence has 41 terms with a value of 41, 34 terms with a value of 8, and the remaining seven 8s can be split into seven pairs, four pairs of $\{7, 1\}$, two pairs of $\{6, 2\}$, and one pair of $\{5, 3\}$, resulting in 14 additional terms). Thus, each $A$-type group can consist of one 41 and one 8, or one 41 and a pair of terms that sum to 8. This results in 41 $A$-type groups, each with a sum of 49. To obtain 49 $B$-type groups with a sum of 41, we can make $x_{1}, x_{2}, \cdots, x_{41}$ each a group, and the remaining numbers can be combined into eight $B$-type groups: four groups of $\{8, 8, 8, 8, 8, 1\}$, two groups of $\{8, 8, 8, 8, 7, 2\}$, one group of $\{8, 8, 8, 8, 6, 3\}$, and one group of $\{8, 8, 7, 7, 6, 5\}$. Therefore, the minimum value of $n$ is 89.