10. (15 points) Given a sequence of positive numbers $\left\{a_{n}\right\}(n \geqslant 0)$ that satisfies $a_{n}=\frac{a_{n-1}}{m a_{n-2}}(n=2,3, \cdots, m$ is a real parameter $)$. If $a_{2009}=\frac{a_{0}}{a_{1}}$, find the value of $m$.
10. Since $a_{2}=\frac{a_{1}}{m a_{0}}, a_{3}=\frac{\frac{a_{1}}{m a_{0}}}{m a_{1}}=\frac{1}{m^{2} a_{0}}$,
$$
\begin{array}{l}
a_{4}=\frac{\frac{1}{m^{2} a_{0}}}{m \frac{a_{1}}{m a_{0}}}=\frac{1}{m^{2} a_{1}}, a_{5}=\frac{\frac{1}{m^{2} a_{1}}}{m \frac{1}{m^{2} a_{0}}}=\frac{a_{0}}{m a_{1}}, \\
a_{6}=\frac{\frac{a_{0}}{m a_{1}}}{m \frac{1}{m^{2} a_{1}}}=a_{0}, a_{7}=\frac{a_{0}}{m \frac{a_{0}}{m a_{1}}}=a_{1},
\end{array}
$$

Therefore, the sequence $\left\{a_{n}\right\}$ is a periodic sequence with a period of 6.
Thus, $\frac{a_{0}}{a_{1}}=a_{2009}=a_{5}=\frac{a_{0}}{m a_{1}} \Rightarrow m=1$.