10. Dividing both sides of the given equation by $n(n+1)$, we get
$$
\begin{array}{l}
\frac{a_{n}^{2}}{n}-\frac{2 a_{n+1}^{2}}{n+1}+\frac{a_{n+1} a_{n}}{\sqrt{n(n+1)}}=0 \\
\Rightarrow\left(\frac{a_{n+1}}{\sqrt{n+1}}-\frac{a_{n}}{\sqrt{n}}\right)\left(\frac{a_{n+1}}{\sqrt{n+1}}+\frac{2 a_{n}}{\sqrt{n}}\right)=0 .
\end{array}
$$

Since $\left\{a_{n}\right\}$ is a sequence of positive terms, we have
$$
\frac{a_{n+1}}{\sqrt{n+1}}=\frac{a_{n}}{\sqrt{n}} \text {. }
$$

Given $a_{1}=1$, then $a_{n}=\sqrt{n}$.
Thus, $z_{k}=\prod_{n=1}^{k}\left(1-\frac{\mathrm{i}}{a_{n}}\right)=\prod_{n=1}^{k}\left(1-\frac{\mathrm{i}}{\sqrt{n}}\right)$
$\Rightarrow z_{2020}=\prod_{n=1}^{2020}\left(1-\frac{\mathrm{i}}{\sqrt{n}}\right)$
$=\prod_{n=1}^{2019}\left(1-\frac{\mathrm{i}}{\sqrt{n}}\right)\left(1-\frac{\mathrm{i}}{\sqrt{2020}}\right)$
$=\left(1-\frac{\mathrm{i}}{\sqrt{2020}}\right) z_{2019}$
$\Rightarrow z_{2020}-z_{2019}=\frac{\mathrm{i}}{\sqrt{2020}} z_{2019}$
$\Rightarrow\left|z_{2020}-z_{2019}\right|=\frac{1}{\sqrt{2020}}\left|z_{2019}\right|$.
Also, $\left|z_{2019}\right|=\left|\prod_{n=1}^{2019}\left(1-\frac{\mathrm{i}}{\sqrt{n}}\right)\right|$
$=\prod_{n=1}^{2019}\left|\left(1-\frac{\mathrm{i}}{\sqrt{n}}\right)\right|=\prod_{n=1}^{2019} \frac{\sqrt{n+1}}{\sqrt{n}}$
$=\sqrt{2020}$,
hence $\left|z_{2020}-z_{2019}\right|=\frac{1}{\sqrt{2020}}\left|z_{2019}\right|=1$.
10. (20 points) Given the positive sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{1}=1, \\
(n+1) a_{n}^{2}-2 n a_{n+1}^{2}+\sqrt{n^{2}+n} a_{n} a_{n+1}=0, \\
z_{n}=\prod_{k=1}^{n}\left(1-\frac{i}{a_{k}}\right) \text { (i is the imaginary unit). }
\end{array}
$$

Find the value of $\left|z_{2019}-z_{2020}\right|$.