Answer: 230

Solution:

Since $f(w) - w = w(w - 20),$ the two legs at $f(z)$ are $$\begin{gathered} f(z) - z = z(z - 20), \\ f(f(z)) - f(z) = f(z) \\ {}\cdot \bigl(f(z) - 20\bigr) \\ = z(z - 19)(z - 20) \\ {}\cdot (z + 1), \end{gathered}$$ using $f(z) = z(z - 19)$ and $f(z) - 20 = (z - 20)(z + 1).$ They are perpendicular exactly when their quotient $(z - 19)(z + 1)$ is purely imaginary and nonzero.

Write $z = x + 11i.$ The real part of $(z - 19)(z + 1) = z^2 - 18z - 19$ is $x^2 - 121 - 18x - 19,$ so we need $x^2 - 18x - 140 = 0,$ giving $x = 9 \pm \sqrt{221}.$ The form $m + \sqrt{n} + 11i$ with $m, n$ positive integers requires $x = 9 + \sqrt{221}.$

Hence $m + n = 9 + 221 = 230.$