15. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{1}=1, a_{2}=\frac{3}{7}, \\
a_{n}=\frac{a_{n-2} a_{n-1}}{2 a_{n-2}-a_{n-1}}\left(n \in \mathbf{Z}_{+}, n \geqslant 3\right) .
\end{array}
$$

If $a_{2019}=\frac{p}{q}\left(p 、 q \in \mathbf{Z}_{+},(p, q)=1\right)$, then $p+q=(\quad)$.
(A) 2020
(B) 4039
(C) 6057
(D) 6061
(E) 8078
15. E.
$$
\begin{array}{l}
\text { Given } a_{n}=\frac{a_{n-2} a_{n-1}}{2 a_{n-2}-a_{n-1}} \\
\Rightarrow \frac{1}{a_{n}}=\frac{2 a_{n-2}-a_{n-1}}{a_{n-2} a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}} \\
\Rightarrow \frac{1}{a_{n}}-\frac{1}{a_{n-1}}=\frac{1}{a_{n-1}}-\frac{1}{a_{n-2}} \\
=\frac{1}{a_{2}}-\frac{1}{a_{1}}=\frac{4}{3} .
\end{array}
$$

Then $\left\{\frac{1}{a_{n}}\right\}$ is an arithmetic sequence with the first term 1 and common difference $\frac{4}{3}$.
$$
\begin{array}{l}
\text { Therefore, } \frac{1}{a_{n}}=1+\frac{4}{3}(n-1) \\
\Rightarrow \frac{1}{a_{2019}}=\frac{8075}{3} \\
\Rightarrow a_{2019}=\frac{3}{8075} \\
\Rightarrow p+q=8078 .
\end{array}
$$