Answer: 330

Solution:

Both cubics have real coefficients, so their non-real roots come in conjugate pairs: the roots of the first are $-20$ and $m \pm \sqrt{n}\,i,$ and the roots of the second are $-21$ and $m \pm \sqrt{n}\,i.$

The first cubic $x^3 + ax + b$ has no $x^2$ term, so its roots sum to $0:$ $-20 + 2m = 0,$ giving $m = 10.$ The second cubic $x^3 + cx^2 + d$ has no $x$ term, so the sum of pairwise products of its roots is $0:$ $$\begin{aligned} &(m + \sqrt{n}\,i)(m - \sqrt{n}\,i) \\ &\quad {}+ (-21)(2m) \\ &= m^2 + n - 42m = 0, \end{aligned}$$ so $n = 420 - 100 = 320.$ Then $m + n = 10 + 320 = 330.$