10. Given a sequence $\left\{a_{n}\right\}$ whose terms are all non-negative real numbers, and satisfies: for any integer $n \geqslant 2$, we have $a_{n+1}=$ $a_{n}-a_{n-1}+n$. If $a_{2} a_{2022}=1$, find the maximum possible value of $a_{1}$.
According to the problem, when $n \geqslant 2$, $\Rightarrow a_{n+2}=a_{n+1}-a_{n}+n+1=-a_{n-1}+2 n+1$,
then $a_{n+3}=-a_{n}+2 n+3 \Rightarrow a_{n+6}=-a_{n+3}+2 n+9=a_{n}+6$.
Also, $a_{3}=a_{2}-a_{1}+2, a_{4}=a_{3}-a_{2}+3=-a_{1}+5, a_{5}=a_{4}-a_{3}+4=-a_{2}+7$,
$a_{6}=a_{5}-a_{4}+5=a_{1}-a_{2}+7 \Rightarrow a_{2022}=a_{6 \times 337}=a_{6}+6 \times 336=2023+a_{1}-a_{2}$,
thus $a_{2} a_{2022}=1 \Rightarrow 2023+a_{1}-a_{2}=\frac{1}{a_{2}} \Rightarrow a_{1}=a_{2}+\frac{1}{a_{2}}-2023$.
From $a_{3}=a_{2}-a_{1}+2=-\frac{1}{a_{2}}+2025 \geqslant 0 \Rightarrow a_{2} \geqslant \frac{1}{2025}$,
$a_{6}=a_{1}-a_{2}+7=\frac{1}{a_{2}}-2016 \geqslant 0 \Rightarrow a_{2} \leqslant \frac{1}{2016} \Rightarrow a_{2} \in\left[\frac{1}{2025}, \frac{1}{2016}\right]$.
Noting that the function $a_{1}=a_{2}+\frac{1}{a_{2}}-2023$ is monotonically decreasing on $a_{2} \in\left[\frac{1}{2025}, \frac{1}{2016}\right]$,
the maximum value of $a_{1}$ is $\frac{1}{2025}+2=\frac{4051}{2025}$.