15. There are 10 students standing in a row, and their birthdays are in different months. There are $n$ teachers who will select these students to join $n$ interest groups. Each student is selected by exactly one teacher, and the order of the students is maintained. Each teacher must select students whose birthdays are in months that are either strictly increasing or strictly decreasing (selecting one or two students is also considered strictly increasing or decreasing). Each teacher should select as many students as possible. For all possible orderings of the students, find the minimum value of $n$.
15. If $n \leqslant 3$, let's assume the birth months of these 10 students are $1,2, \cdots, 10$.

When the students are sorted by their birthdays as $4,3,2,1,7,6,5,9, 8,10$, there exists at least one teacher who must select two students from the first four. Since the birth months of these two students are decreasing, and the birth months of the last six students are all greater than those of the first four, this teacher cannot select any of the last six students; among the remaining no more than two teachers, there must be one who has to select two students from the fifth to the seventh. Similarly, this teacher cannot select any of the last three students; the remaining no more than one teacher also cannot select any of the last three students, leading to a contradiction.

Below is the proof: For any distinct ordered sequence of real numbers $a_{1}, a_{2}, \cdots, a_{m}$, when $m \geqslant 5$, there must exist three numbers $a_{i}, a_{j}, a_{k}(i<a_{j}>a_{k}$.
Let the maximum and minimum numbers be $a_{s}$ and $a_{t}$. Without loss of generality, assume $s<a_{s+1}>a_{t} ; s+1=t$.
Since $m \geqslant 5$, there must be at least two numbers either before $a_{s}, a_{s+1}$ or after $a_{s}, a_{s+1}$.

Without loss of generality, assume there are two numbers $a_{s+2}$ and $a_{s+3}$ after $a_{s}, a_{s+1}$. Thus,
$$
a_{s}>a_{s+2}>a_{s+3} \text { or } a_{s+1}<a_{s+2}<a_{s+3}
$$

must hold.
Using the above conclusion, when $n=4$, the first teacher can select at least 3 students; if the remaining students are greater than or equal to 5, the second teacher can also select at least 3 students; at this point, the number of remaining students does not exceed 4, and can be selected by the two teachers.
Therefore, the minimum value of $n$ is 4.