Without loss of generality, we can assume that $a_{1} \leq a_{2} \leq \cdots \leq a_{2 n}$. We distinguish two cases:

- If $a_{n}+a_{2 n-1} \geq 0$ then we have $a_{i}+a_{2 n-1} \geq 0$ for $i=n, \cdots, 2 n-2$, and $a_{i}+a_{2 n} \geq 0$ for $i=n \cdots, 2 n-1$. This provides at least $2 n-1$ non-negative sums.
- If $a_{n}+a_{2 n-1} < 0$
$$0 > a_{n}+a_{2 n-1} \geq a_{n-1}+a_{2 n-2} \geq \cdots \geq a_{2}+a_{n+1},$$
then
$$a_{2}+a_{3}+\cdots+a_{n-1}+a_{n+1}+\cdots+a_{2 n-2} < 0.$$
Let $\ell$ be the number of non-negative terms among $a_{1}, \ldots, a_{2 n}$. Then there are at least $\frac{\ell(\ell-1)}{2} \geq \frac{n(n+1)}{2}$ pairs $\left(a_{i}, a_{j}\right)$ with $i<j, a_{i} \geqslant 0$ and $a_{j} \geqslant 0$. Since $\frac{n(n+1)}{2}-(2 n-1)=\frac{n^{2}-3 n+2}{2}=\frac{(n-1)(n-2)}{2} \geqslant 0$, there are at least $2 n-1$ pairs $\left(a_{i}, a_{j}\right)$ with $i<j$ such that $a_{i}+a_{j} \geqslant 0$.
First case: $\ell > n$. Then there are at least $\frac{\ell(\ell-1)}{2} \geq \frac{n(n+1)}{2}$ pairs $\left(a_{i}, a_{j}\right)$ with $i<j, a_{i} \geqslant 0$ and $a_{j} \geqslant 0$. Since $\frac{n(n+1)}{2}-(2 n-1)=\frac{n^{2}-3 n+2}{2}=\frac{(n-1)(n-2)}{2} \geqslant 0$, there are at least $2 n-1$ pairs $\left(a_{i}, a_{j}\right)$ with $i<j$ such that $a_{i}+a_{j} \geqslant 0$.
Second case: $\ell \leqslant n$. Let $c_{1} \leqslant \cdots \leqslant c_{\ell}$ be the smallest integers among $a_{1}, \ldots, a_{2 n}$. Since $\ell \leqslant n$, we have $c_{\ell}<0$. Moreover, $\sum_{i=1}^{2 n} a_{i}$ is equal to the sum of $\sum_{i=1}^{\ell}\left(b_{i}+c_{i}\right)$ and negative terms, so $\sum_{i=1}^{\ell}\left(b_{i}+c_{i}\right) \geqslant 0$. Since $b_{\ell}+c_{\ell} \geqslant b_{i}+c_{i}$ for all $i$, we have $b_{\ell}+c_{\ell} \geqslant 0$.
Thus, we already have $2 n-\ell$ pairs $\left(a_{i}, a_{j}\right)$ by taking $a_{j}=b_{\ell}$ and $a_{i}$ other than $c_{1}, \ldots, c_{\ell-1}, b_{\ell}$.
Furthermore, for all $k=1, \ldots, \ell-1$, we have $\sum_{i=1}^{\ell}\left(b_{i}+c_{i+k}\right) \geqslant 0$ (where by convention $c_{\ell+1}=c_{1}$, $c_{\ell+2}=c_{2}$, etc.), so for all $k$ there exists $i$ such that $b_{i}+c_{i+k} \geqslant 0$. This provides another $\ell-1$ pairs.