Originally, there are $ (n-2)(n+8) = n^2 + 6n - 16 $ members in the band. In the second formation, there are at least $ 4 $ more than $ (n)(2n-3) = 2n^2 - 3n $ members in the band. Thus, $ n^2 + 6n - 16 \ge 2n^2 - 3n + 4 $, or simplifying, $$0 \ge n^2 - 9n + 20.$$ The quadratic expression factors as $ 0 \ge (n-4)(n-5) $. Thus $ 4 \le n \le 5 $, and $ n = 4,5 $. We can verify that both values work, from which it follows that the answer is $ 4+5 = \boxed{9} $.