Four, (50 points) For $n$ distinct positive integers, among any six numbers, there are at least two numbers such that one can divide the other. Find the minimum value of $n$ such that among these $n$ numbers, there must exist six numbers where one can be divided by the other five.


The smallest positive integer $n=26$.
The proof is divided into two steps.
【Step 1】When $n \leqslant 25$, the condition is not satisfied.
Construct the following 25 positive integers:
(1) $2^{5}, 2^{4}, 2^{3}, 2^{2}, 2^{1}$;
(2) $3^{5}, 3^{4}, 3^{3}, 3^{2}, 3^{1}$;
(3) $5^{5}, 5^{4}, 5^{3}, 5^{2}, 5^{1}$;
(4) $7^{5}, 7^{4}, 7^{3}, 7^{2}, 7^{1}$;
(5) $11^{5}, 11^{4}, 11^{3}, 11^{2}, 11^{1}$.

On one hand, divide these 25 positive integers into five groups. Then, any selection of six numbers will definitely have two numbers in the same group. Clearly, one of these two numbers in the same group can divide the other.

On the other hand, since each group has only 5 numbers, the six selected numbers must come from at least two groups, meaning that there is no number among the six selected numbers that can be divided by the other five.
Therefore, when $n=25$, the condition is not satisfied.
For $n<25$, a similar proof can be made.
【Step 2】When $n=26$, the condition is satisfied.
If a set of numbers is all within the given 26 positive integers, denote the largest one as $a$. Among the remaining 25 numbers, there are no multiples of $a$, and all divisors of $a$ are included in this set, then this set is called a "good set" (a good set can contain only one number).

Next, we prove: there can be at most five good sets. Otherwise, there must be six good sets.

Consider the largest numbers in these six good sets, denoted as $a, b, c, d, e, f$. By the condition, among these six numbers, there must be one that can divide another, say $b \mid a$, meaning that the divisor $b$ of $a$ is not in the good set containing $a$.

This contradicts the definition of a good set, hence there can be at most five good sets.

Since there can be at most five good sets, and there are 26 given positive integers, there must be at least one good set with six numbers.

Consider the largest number in this good set. By the definition of a good set, there are at least five other numbers in this set that can divide this largest number.
In conclusion, the smallest positive integer $n=26$.