N numbers, N/2 pairs. Minimizing the maximum sum of a pairing. Proving greedy algorithm
Can you see why $\max((-2)+5, 3+4) \lt \max(-2+3, 4+5)$?

The reason is simple. Because on the right hand side, the maximum number 5 is not  paired with the minimum number.



Let the numbers are $a_1\le a_2\le \cdots\le a_n$. Let the numbers be paired in some way. 


If $a_n$ is paired with $a_1$, we are done at this round.
Suppose $a_n$ is paired with $a_j$, $j\not= 1$. Then $a_1$ is paired with $a_k$ for some $k\not= n$. So we have two pairs, $\{a_n, a_j\}$ and $\{a_1, a_k\}$. The sums of these two pairs are $a_n + a_j$ and $a_1 + a_k$, the large one of which is $a_n+a_j$.

Let us switch $a_j$ and $a_1$ so that $a_n$ will pair with $a_1$, and $a_j$ will pair with $a_k$. The sums of the two new pairs are $a_n + a_1$ and $a_j + a_k$, both of which is at most $a_n+a_j$, i.e., the large one of them is at most $a_n+a_j$. So after the switch, the maximum sum of the pairs involving $a_n, a_j, a_k, a_1$ does not increase. Since other pairs stays the same, So after the switch, the maximum sum of all pairs does not increase.


Continuing this process, we will make sure the largest number remaining be paired with the smallest number remaining at each round. The maximum sum of the pairs will never increase at each round. After $n/2$ rounds, we will reach the pairing where $a_k$ is paired with $a_{n+1-k}$.

You can see the above approach is indeed the "greedy stays ahead" approach.