15. The sequence of positive integers $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{2}=15, \\
2+\frac{4}{a_{n}+1}<\frac{a_{n}}{a_{n}-4 n+2}+\frac{a_{n}}{a_{n+1}-4 n-2} \\
<2+\frac{4}{a_{n}-1} .
\end{array}
$$

Calculate: $S_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}$.
15. First, find the general term $a_{n}$.

When $n=1$, the given condition becomes
$$
2+\frac{4}{a_{1}+1}1, \\
\frac{2}{a_{1}-2}+\frac{a_{1}}{9}-\frac{4}{a_{1}-1}0, \text{ i.e., } \left(a_{1}-10\right)\left(a_{1}^{2}-11\right)>2.
$$
Thus, $a_{1} \leqslant 3$ or $a_{1} \geqslant 11$.
From equation (3), we have $a_{1}^{3}-12 a_{1}^{2}+11 a_{1}+36168 \text {. }$

Therefore, $3 \leqslant a_{1} \leqslant 10$.
From equations (4) and (5), we get $a_{1}=3$.
When $n=2$, the given condition becomes
$$
2+\frac{1}{4}<\frac{15}{9}+\frac{15}{a_{3}-10}<2+\frac{2}{7} \text {, }
$$

i.e., $\frac{7}{12}<\frac{15}{a_{3}-10}<\frac{13}{21}$.
Thus, $\frac{315}{13}<a_{3}-10<\frac{180}{7}$, i.e.,
$34+\frac{3}{13}<a_{3}<35+\frac{5}{7}$.
Therefore, $a_{3}=35$.
Notice that $a_{1}+1=4=2^{2}, a_{2}+1=16=4^{2}$, $a_{3}+1=36=6^{2}$.
Next, we prove:
$a_{n}=(2 n)^{2}-1$.
Equation (6) holds for $n=1,2,3$. Assume equation (6) holds for $n(n \geqslant 3)$, consider the case for $n+1$.
From the given condition,
$$
2+\frac{1}{n^{2}}<\frac{2 n+1}{2 n-1}+\frac{4 n^{2}-1}{a_{n+1}-4 n-2}<2+\frac{2}{2 n^{2}-1} \text {, }
$$

i.e., $1+\frac{1}{n^{2}}-\frac{2}{2 n-1}<\frac{4 n^{2}-1}{a_{n+1}-4 n-2}$
$$
\begin{array}{l}
<1+\frac{2}{2 n^{2}-1}-\frac{2}{2 n-1}, \\
\text { also, } \frac{2 n^{3}-3 n^{2}+2 n-1}{n^{2}(2 n-1)}<\frac{(2 n+1)(2 n-1)}{a_{n+1}-2(2 n+1)} \\
<\frac{4 n^{3}-6 n^{2}+2 n+1}{(2 n-1)\left(2 n^{2}-1\right)} \cdots \\
\text { Hence, } \frac{(2 n-1)^{2}\left(2 n^{2}-1\right)}{4 n^{3}-6 n^{2}+2 n+1}<\frac{a_{n+1}}{2 n+1}-2 \\
<\frac{n^{2}(2 n-1)^{2}}{2 n^{3}-3 n^{2}+2 n-1}, \\
\text { i.e., } \frac{(2 n+1)\left(4 n^{3}-6 n^{2}+2 n+1\right)-2}{4 n^{3}-6 n^{2}+2 n+1} \\
\quad<\frac{a_{n+1}}{2 n+1}-2 \\
<\frac{(2 n+1)\left(2 n^{3}-3 n^{2}+2 n-1\right)+1}{2 n^{3}-3 n^{2}+2 n-1},
\end{array}
$$

i.e., $(2 n+1)-\frac{2}{4 n^{3}-6 n^{2}+2 n+1}$
$$
\begin{array}{c}
<\frac{a_{n+1}}{2 n+1}-2 \\
<(2 n+1)+\frac{1}{2 n^{3}-3 n^{2}+2 n-1} .
\end{array}
$$

Thus, $(2 n+1)^{2}+2(2 n+1)-$
$$
\begin{array}{c}
\frac{2(2 n+1)}{4 n^{3}-6 n^{2}+2 n+1} \\
<a_{n+1} \\
<(2 n+1)^{2}+2(2 n+1)+\frac{2 n+1}{2 n^{3}-3 n^{2}+2 n-1} .
\end{array}
$$

Therefore, $(2 n+1)^{2}+2(2 n+1)-1<a_{n+1}$
$$
<(2 n+1)^{2}+2(2 n+1)+1 \text {. }
$$

Since $a_{n+1}$ is an integer, we have
$$
a_{n+1}=(2 n+1)^{2}+2(2 n+1)=[2(n+1)]^{2}-1 \text {. }
$$

Hence, by induction, equation (6) holds for any positive integer $n$, i.e., $a_{n}=(2 n)^{2}-1$.
Next, calculate $S_{n}$. Notice
$$
\frac{1}{a_{n}}=\frac{1}{(2 n-1)(2 n+1)}=\frac{1}{2}\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right) \text {, }
$$

then $S_{n}=\sum_{k=1}^{n} \frac{1}{a_{k}}=\frac{1}{2} \sum_{k=1}^{n}\left(\frac{1}{2 k-1}-\frac{1}{2 k+1}\right)$
$$
=\frac{1}{2}\left(1-\frac{1}{2 n+1}\right)=\frac{n}{2 n+1} .
$$