Assume, without loss of generality, that
$$0<a_{1} \leq a_{2} \leq \cdots \leq a_{n} \leq \frac{1}{2} \leq a_{2 n} \leq a_{2 n-1} \leq \cdots \leq a_{n+1}<1,$$
and let
$$K=\min \left\{\left|a_{i}^{2}-a_{j}^{2}\right|, \text{ with } i, j=1, \ldots, 2 n, \text{ and } i \neq j\right\} .$$
We have
$$K \leq a_{2 n}^{2}-a_{n}^{2}=\left(a_{2 n}+a_{n}\right)\left(a_{2 n}-a_{n}\right)=1-2 a_{n},$$
which means that
$$\frac{1-K}{2} \geq a_{n} .$$
We deduce that
$$\frac{(1-K)^{2}}{4} \geq a_{n}^{2}>a_{n}^{2}-a_{1}^{2}=\sum_{i=1}^{n-1}\left(a_{i+1}^{2}-a_{i}^{2}\right) \geq(n-1) K .$$
This implies that
$$K^{2}-2(2 n-1) K+1>0 .$$
Because $K<1$, we deduce that
$$K<2 n-1-\sqrt{(2 n-1)^{2}-1}=(\sqrt{n}-\sqrt{n-1})^{2},$$
and therefore
$$\sqrt{K}<\sqrt{n}-\sqrt{n-1}=\frac{1}{\sqrt{n}+\sqrt{n-1}} .$$