Solution:

Write $\operatorname{cis}\theta = \cos\theta + i\sin\theta.$ The equation says $z^8(z^{20} - 1) = 1.$ Taking absolute values and using $|z| = 1$ gives $|z^{20} - 1| = 1,$ so $z^{20}$ is at distance $1$ from both $0$ and $1:$ it is $\operatorname{cis}(\pm 60^\circ).$

If $z^{20} = \operatorname{cis} 60^\circ,$ then $z^{20} - 1 = \operatorname{cis} 120^\circ,$ so $z^8 = \operatorname{cis}(-120^\circ)$ and $$z^4 = \frac{z^{20}}{(z^8)^2} = \operatorname{cis}(60^\circ + 240^\circ) = \operatorname{cis} 300^\circ,$$ which means $4\theta \equiv 300^\circ,$ i.e. $\theta \equiv 75^\circ \pmod{90^\circ}.$ Conversely every such $\theta$ works, since then $z^{20} = (z^4)^5 = \operatorname{cis} 60^\circ$ and $z^8 = (z^4)^2 = \operatorname{cis}(-120^\circ).$ The case $z^{20} = \operatorname{cis}(-60^\circ)$ similarly gives exactly $\theta \equiv 15^\circ \pmod{90^\circ}.$

So the $2n = 8$ angles in increasing order are $15, 75, 105, 165, 195, 255, 285, 345,$ and $\theta_2 + \theta_4 + \theta_6 + \theta_8 = 75 + 165 + 255 + 345 = 840.$