14. Given the sequence $\left\{a_{n}\right\}(n \geqslant 0)$ satisfies $a_{0}=0$, $a_{1}=1$, for all positive integers $n$, we have
$$
a_{n+1}=2 a_{n}+2007 a_{n-1} \text {. }
$$

Find the smallest positive integer $n$ such that $2008 \mid a_{n}$.
14. Solution 1: Let $m=2008, a_{n+1}=2 a_{n}+2007 a_{n-1}$, the characteristic equation is $\lambda^{2}-2 \lambda-2007=0$, with characteristic roots $1 \pm \sqrt{m}$. Combining $a_{0}=0, a_{1}=1$, we get
$$
a_{n}=\frac{1}{2 \sqrt{m}}\left[(1+\sqrt{m})^{n}-(1-\sqrt{m})^{n}\right] \text {. }
$$

By the binomial theorem,
$$
a_{n}=\frac{1}{2 \sqrt{m}}\left[\sum_{k=0}^{n} \mathrm{C}_{n}^{k} m^{\frac{k}{2}}-\sum_{k=0}^{n}(-1)^{k} \mathrm{C}_{n}^{k} m^{\frac{k}{2}}\right] \text {. }
$$

When $n$ is odd,
$$
a_{n}=\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{3} m+\cdots+\mathrm{C}_{n}^{n-2} m^{\frac{n-3}{2}}+\mathrm{C}_{n}^{n} m^{\frac{n-1}{2}} \text {; }
$$

When $n$ is even,
$$
a_{n}=\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{3} m+\cdots+\mathrm{C}_{n}^{n-3} m^{\frac{n-4}{2}}+\mathrm{C}_{n}^{n-1} m^{\frac{n-2}{2}} \text {. }
$$

Thus, $m\left|a_{n} \Leftrightarrow m\right| \mathrm{C}_{n}^{1}$, i.e., $2008 \mid n$.
Therefore, the smallest positive integer that satisfies the condition is 2008.
Solution 2: The following are all $a_{n}$ modulo 2008, then $a_{n+1} \equiv 2 a_{n}-a_{n-1}$, i.e.,
$$
a_{n+1}-a_{n} \equiv a_{n}-a_{n-1} .
$$

Therefore, the sequence $\left\{a_{n}\right\}$ has the characteristics of an arithmetic sequence modulo 2008.
Since $a_{0}=0, a_{1}=1$, we have $a_{n} \equiv n$.
Thus, $2008 \mid n$.
Therefore, the smallest positive integer that satisfies the condition is 2008.