10. (20 points) Given the sequence $\left\{a_{n}\right\}_{n \geqslant 0}$ satisfies $a_{0}=0$, $a_{1}=1$, and for all positive integers $n$,
$$
a_{n+1}=2 a_{n}+2013 a_{n-1} \text {. }
$$

Find the smallest positive integer $n$ such that $2014 \mid a_{n}$.
10. Below are $a_{n}$ modulo 2014.

Then $a_{n+1} \equiv 2 a_{n}-a_{n-1} \Rightarrow a_{n+1}-a_{n} \equiv a_{n}-a_{n-1}$.
Therefore, the sequence $\left\{a_{n}\right\}$ has the characteristics of an arithmetic sequence modulo 2014.
Since $a_{0}=0, a_{1}=1$, we have $a_{n} \equiv n$.
Thus, the smallest positive integer $n$ such that $2014 \mid a_{n}$ is 2014.