15. For any positive integer $m$, the set
$$
\{m, m+1, m+2, \cdots, m+99\}
$$

has the property that in any $n(n \geqslant 3)$-element subset, there are always three elements that are pairwise coprime. Find the minimum value of $n$.
15. Consider the set $\{1,2, \cdots, 100\}$ when $m=1$, and its 67-element subset, whose elements are even numbers and odd numbers divisible by 3, i.e.,
$$
P=\{2,4, \cdots, 100,3,9, \cdots, 99\} .
$$

Clearly, there do not exist three pairwise coprime elements in set $P$.
Thus, $n \leqslant 67$ does not meet the requirement.
We now prove that $n=68$ meets the requirement.
First, we prove a lemma.
Lemma For any positive integer $m$, in any five-element subset of the set
$$
\{m, m+1, \cdots, m+5\}
$$

there are always three elements that are pairwise coprime.
Proof Let $A$ be a five-element subset of the set $\{m, m+1, \cdots, m+5\}$.

Notice that $m, m+1, \cdots, m+5$ consist of three odd and three even numbers, and exactly one of them is a multiple of 5.

Thus, if set $A$ contains three odd numbers, these three odd numbers must be pairwise coprime, and the conclusion holds.

If set $A$ contains two odd and three even numbers, since at most one of the three even numbers is a multiple of 3, and at most one is a multiple of 5, there must be one even number that is neither a multiple of 3 nor 5. This number is coprime with the two odd numbers, and the conclusion holds.
Returning to the original problem.
For any positive integer $m$, partition the set
$$
\{m, m+1, \cdots, m+99\}
$$

into the following 17 sets:
$$
A_{i}=\{m+6(i-1)+j \mid j=0,1, \cdots, 5\}
$$
$(i=1,2, \cdots, 16)$,
$$
A_{17}=\{m+96, m+97, m+98, m+99\} .
$$

Let $M$ be a 68-element subset of the set $\{m, m+1, \cdots, m+99\}$.
(1) If set $M$ has four elements from set $A_{17}$, since when $m$ is odd, $m+96, m+97, m+98$ are pairwise coprime; and when $m$ is even, $m+97, m+98, m+99$ are pairwise coprime, therefore, set $M$ contains at least three elements that are pairwise coprime.
(2) If set $M$ has at most three elements from set $A_{17}$, then set $M$ has at least 65 elements from sets $A_{1}, A_{2}, \cdots, A_{16}$.

By the pigeonhole principle, set $M$ must have at least five elements from one of the sets, say $A_{1}$. By the lemma, there exist three pairwise coprime elements among them. Therefore, set $M$ always contains three pairwise coprime elements.

Thus, $n=68$ meets the requirement, i.e., for any positive integer $m$, any 68-element subset of the set $\{m, m+1, \cdots, m+99\}$ contains three elements that are pairwise coprime.
In summary, the minimum value of $n$ is 68.