Answer: 784

Solution:

The zeroes are $z_j = 8\left(\cos\frac{\pi j}{6} + i\sin\frac{\pi j}{6}\right)$ for $j = 1, \ldots, 12,$ and $\operatorname{Re}(iz_j) = -\operatorname{Im}(z_j).$ Since the choices are independent, the maximum real part of the sum is $\sum_j 8\max\left(\cos\frac{\pi j}{6},\, -\sin\frac{\pi j}{6}\right).$

Comparing the two values, $-\sin\frac{\pi j}{6}$ is larger exactly for $j = 5, \ldots, 10.$ The cosines kept, for $j = 1, 2, 3, 4, 11, 12,$ sum to $$\begin{aligned} &\frac{\sqrt{3}}{2} + \frac{1}{2} + 0 - \frac{1}{2} \\ &\quad {}+ \frac{\sqrt{3}}{2} + 1 = 1 + \sqrt{3}, \end{aligned}$$ and the values $-\sin\frac{\pi j}{6}$ kept, for $j = 5, \ldots, 10,$ sum to $$\begin{aligned} &-\frac{1}{2} + 0 + \frac{1}{2} + \frac{\sqrt{3}}{2} \\ &\quad {}+ 1 + \frac{\sqrt{3}}{2} = 1 + \sqrt{3}. \end{aligned}$$

The maximum is $$\begin{aligned} &8\left(2 + 2\sqrt{3}\right) \\ &= 16 + 16\sqrt{3} \\ &= 16 + \sqrt{768}, \end{aligned}$$ so $m + n = 16 + 768 = 784.$