jmc

From $z^{24}=1,$ $z^{24}-1=0,$ so $$(z^{12}+1)(z^{12}-1)=0.$$Then $$(z^{12}+1)(z^6+1)(z^6-1)=0.$$Thus, for 6 of the roots, $z^6=-1,$ for another 6 of the roots, $z^6=1,$ and for the remaining 12 roots, $(z^6{)}^2+1=0,$ so $z^6$ is not real. Therefore, for $\boxed{12}$ of the roots, $z^6$ is real.