1. Given the sequence $\left\{x_{n}\right\}$ satisfies $x_{n}=\frac{n}{n+2016}$. If $x_{2016}=x_{m} x_{n}$, then a solution for the positive integers $m 、 n$ is $\{m, n\}$ $=$ . $\qquad$
1. $\{4032,6048\}$. (Not unique)
$$
\begin{array}{l}
\text { Given } x_{2016}=x_{m} x_{n} \\
\Rightarrow \frac{1}{2}=\frac{m}{m+2016} \times \frac{n}{n+2016} \\
\Rightarrow(m+2016)(n+2016)=2 m n \\
\Rightarrow(m-2016)(n-2016)=2 \times 2016^{2} \text {. } \\
\text { If } m-2016=2016, n-2016=2 \times 2016 \\
\Rightarrow m=4032, n=6048 .
\end{array}
$$