9. Mathematician Fibonacci discovered a sequence of numbers while studying the rabbit reproduction problem: $1,1,2,3,5,8,13, \cdots$ The characteristic of this sequence is: the first two numbers are both 1, and from the third number onwards, each number is equal to the sum of the two preceding numbers. The sequence $\left\{a_{n}\right\}$ formed by such a series of numbers is called the “Fibonacci sequence”. Then
$$
\begin{array}{l}
\left(a_{1} a_{3}+a_{2} a_{4}+a_{3} a_{5}+\cdots+a_{2019} a_{2021}\right)- \\
\left(a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+\cdots+a_{2020}^{2}\right) \\
=
\end{array}
$$
9.1.

From the problem, we know
$$
\begin{array}{l}
a_{1} a_{3}-a_{2}^{2}=1, a_{2} a_{4}-a_{3}^{2}=-1, \\
a_{3} a_{5}-a_{4}^{2}=1, a_{4} a_{6}-a_{5}^{2}=-1, \\
\cdots \cdots
\end{array}
$$

From this, we can deduce the conclusion
$$
a_{n} a_{n+2}-a_{n+1}^{2}=(-1)^{n+1}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Thus, the original expression is
$$
\begin{aligned}
= & \left(a_{1} a_{3}-a_{2}^{2}\right)+\left(a_{2} a_{4}-a_{3}^{2}\right)+ \\
& \left(a_{3} a_{5}-a_{4}^{2}\right)+\cdots+\left(a_{2019} a_{2021}-a_{2020}^{2}\right),
\end{aligned}
$$

which is the sum of 1009 -1s and 1010 1s, resulting in 1.