For example, $5 n$ positive integers $x_{1}, x_{2}, \cdots, x_{n}$ have a sum of 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 $n$.


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 A-type groups.

If two groups $A_{i}$ and $B_{j}$ have two common terms $x_{i}$ 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.

Construct a graph model: Let the points $u_{1}, u_{2}, \cdots, u_{41}$ represent the groups $A_{1}, A_{2}, \cdots, A_{41}$, and the points $v_{1}, v_{2}, \cdots, v_{49}$ represent the groups $B_{1}, B_{2}, \cdots, B_{49}$. 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.
Next, we prove that the graph $G$ is connected.
If the largest connected component of the graph $G$ is $G^{\prime}$, and the number of vertices in the component $G^{\prime}$ is less than 90, let there be $a$ A-type vertices $u_{k_{1}}, u_{k_{2}}, \cdots, u_{k_{4}}$ and $b$ B-type vertices $v_{x_{1}}, v_{x_{2}}, \cdots, v_{x_{1}}$ in the component $G^{\prime}$, where $a+b<90$. Then, in the corresponding A-type groups $A_{k_{1}}, A_{k_{2}}, \cdots, A_{k_{n}}$ and B-type groups $B_{s_{1}}, B_{s_{2}}, \cdots, B_{s_{b}}$, each number $x_{i}$ in the A-type group $A_{k_{i}}$ must appear in some B-type group $B_{x_{j}}$, and each number $x_{j}$ in the B-type group $B_{x_{j}}$ must appear in some A-type group $A_{r_{i}}$ (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 $A_{k_{1}}, A_{k_{2}}, \cdots, A_{k_{1}}$ should equal the sum of the numbers in the $b$ B-type groups $B_{x_{1}}, B_{s_{2}}, \cdots, B_{s_{b}}$, i.e., $49a = 41b$, from which it follows that $41 \mid a$ and $49 \mid b$.
Thus, $a + b \geq 41 + 49 = 90$, which is a contradiction.
Therefore, the graph $G$ is connected.
Thus, the graph $G$ has at least $90 - 1 = 89$ edges, i.e., $n \geq 89$.
On the other hand, we can construct a sequence of 89 terms $x_{1}, x_{2}, \cdots, x_{89}$ that satisfies the conditions of the problem.
For example, take
\[
\begin{array}{l}
x_{1} = x_{2} = \cdots = x_{41} = 41, \\
x_{42} = x_{43} = \cdots = x_{75} = 8, \\
x_{76} = x_{77} = x_{78} = x_{79} = 7, \\
x_{80} = x_{81} = x_{82} = x_{83} = 1, \\
x_{84} = x_{85} = 6, \\
x_{86} = x_{87} = 2, x_{88} = 5, x_{89} = 3,
\end{array}
\]
i.e., the 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 $\{5,3\}$, yielding 14 more terms.

Thus, each A-type group can consist of one 41, one 8, or one 41 and a pair of terms that sum to 8, resulting in 41 A-type groups, each with a sum of 49.

To obtain 49 B-type groups with a sum of 41, we can form one group for each of $x_{1}, x_{2}, \cdots, x_{41}$, 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 $\{8,8,7,7,6,5\}$.
Thus, the minimum value of $n$ is 89.