Example 5 Choose $n$ numbers from $1,2, \cdots, 9$, among which there must be several numbers (at least one, or all), the sum of which can be divisible by 10. Find the minimum value of $n$.
(2008, National Junior High School Mathematics Competition)
Solution: When $n \leqslant 4$, since no combination of $n$ numbers from $1,3,5,8$ can sum to a multiple of 10, we have $n \geqslant 5$.

Below, we prove that $n=5$ meets the requirement, i.e., we prove that from $1,2, \cdots, 9$, any 5 numbers chosen will always include some numbers whose sum is a multiple of 10.
We use proof by contradiction.
Assume that 5 numbers can be chosen such that no combination of these numbers sums to a multiple of 10. Divide $1,2, \cdots, 9$ into the following 5 groups:
$$
\{1,9\},\{2,8\},\{3,7\},\{4,6\},\{5\},
$$

Then the chosen numbers can only contain one number from each group, thus 5 must be chosen.
(1) If 1 is chosen from the first group, then 9 is not chosen. Since $1+5+4=10$, 4 is not chosen, thus 6 is chosen. Since $1+6+3=10$, 3 is not chosen, thus 7 is chosen. Since $1+7+2=10$, 2 is not chosen, thus 8 is chosen. But $5+7+8=20$ is a multiple of 10, a contradiction. (2) If 9 is chosen from the first group, then 1 is not chosen. Since $9+5+6=20$, 6 is not chosen, thus 4 is chosen. Since $9+4+7=20$, 7 is not chosen, thus 3 is chosen. Since $9+3+8=20$, 8 is not chosen, thus 2 is chosen. But $5+3+2=10$ is a multiple of 10, a contradiction. In summary, the minimum value of $n$ is 5.