Let $m$ and $n$ be integers such that all the roots of the equation $[(x^{2} + mx + 20)(x^{2} + 17x + n)] = 0$ are negative integers. The smallest possible value of $(m + n)$ is :

For roots to be negative integers, x^2 + mx + 20 = (x+a)(x+b) where a,b > 0 and ab=20. Possible pairs (a,b): (1,20), (2,10), (4,5). m = a+b: 21, 12, 9. For x^2 + 17x + n, roots c,d > 0 and c+d=17, cd=n. Pairs (c,d): (1,16), (2,15), (3,14), (4,13), (5,12), (6,11), (7,10), (8,9). n = cd: 16, 30, 42, 52, 60, 66, 70, 72. Min m+n = 9+16 = 25.