15. If $P$ is a $2n$-digit positive integer composed of the digits $1,2,3,4,5,6,7, 8,9$, and simultaneously satisfies the following two conditions:
(1) The digits $1,2, \cdots, n$ each appear twice in $P$;
(2) Between each pair of the same digit $i(i=1,2, \cdots, n)$, there are exactly $i$ digits.

In this case, we call such a positive integer $P$ a "good number". For example, when $n=3$, $P$ can be 312132.

Determine the value of the positive integer $n$ that satisfies the conditions, and write down a corresponding good number $P$ for each.
15. From the definition of a good number, we know that $n \leqslant 9$.

For a good number $P$, considering the positions of the digits from left to right, if the first occurrence of the digit $i (i=1,2, \cdots, n)$ is denoted as $a_{i}$, then according to the problem, the second occurrence of the digit $i (i=1,2, \cdots, n)$ should be at position $a_{i}+(i+1)$. Therefore,
$$
\sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n}\left[a_{i}+(i+1)\right]=\sum_{k=1}^{2 n} k .
$$

Let $S=\sum_{i=1}^{n} a_{i}$, then
$$
S+S+\frac{n[2+(n+1)]}{2}=\frac{2 n(1+2 n)}{2},
$$

which simplifies to $S=\frac{n(3 n-1)}{4}$.
Since $S$ is a positive integer, $n(3 n-1)$ must be divisible by 4.
Also, since $n$ is a positive integer, we have
$n=3$ or 4 or 7 or 8.
When $n=3$, it is already given in the problem;
When $n=4$, a good number $P$ can be 41312432;
When $n=7$, a good number $P$ can be 71316435724625;
When $n=8$, a good number $P$ can be
8131573468524726.
(Li Guomin provided)