Given 2 integer vectores (add zeros to shortest if necessary) we can sum them term by term to get a new integer vector.
$$ (1,2,3) , (1,5,7) $$
If we add the possibility to permute components from second one before addition, then we get $n!$ possible results.
$$ (2,7,10) , (6,3,10) , (2,9,8) , (8,3,8) , (6,9,4) , (8,7,4) $$
I am interested in getting upper bound on minimal component from each vector.
$$max\{2,3,2,3,4,4\} = 4$$
Call the zero-padded vectors $u$ and $v$, so in your example $u = (1,2,3)$ and $v = (1,5,7)$.
Observe that $\max_i(u_i) + \min_j(v_j)$ is an upper bound on your max-min: in every permutation, $\min_j(v_j)$ must be paired with some $u_i$, and that paired element can't be greater than $\max_i(u_i)$.
Similarly, $\min_i(u_i) + \max_j(v_j)$ is an upper bound.
Note that we can't get a tight upper bound just by considering the extrema of the two sets: $u=v=(1,4,64)$ has max-min $8$ when we pair the $1$s with the $64$s.
