jmc

For $n\ne 0,$ we can write $$1+z^n+z^{2n}=\frac{z^{3n}-1}{z^n-1},$$so $$\frac{1}{1+z^n+z^{2n}}=\frac{z^n-1}{z^{3n}-1}.$$Since $z^{23}=1,$ $z^{23n}=1,$ so $z^n=z^{24n}.$ Hence, $$\frac{z^n-1}{z^{3n}-1}=\frac{z^{24n}-1}{z^{3n}-1}=1+z^{3n}+z^{6n}+\cdots +z^{21n}.$$Then $$\sum _{n=0}^{22}\frac{1}{1+z^n+z^{2n}}=\frac{1}{3}+\sum _{n=1}^{22}\frac{1}{1+z^n+z^{2n}},$$and $$\begin{array}{rl}\sum _{n=1}^{22}\frac{1}{1+z^n+z^{2n}}& =\sum _{n=1}^{22}(1+z^{3n}+z^{6n}+\cdots +z^{21n})\\ & =\sum _{n=1}^{22}\sum _{m=0}^7{z}^{3mn}\\ & =\sum _{m=0}^7\sum _{n=1}^{22}{z}^{3mn}\\ & =22+\sum _{m=1}^7\sum _{n=1}^{22}{z}^{3mn}\\ & =22+\sum _{m=1}^7(z^{3m}+z^{6m}+z^{9m}+\cdots +z^{66m})\\ & =22+\sum _{m=1}^7{z}^{3m}(1+z^{3m}+z^{6m}+\cdots +z^{63m})\\ & =22+\sum _{m=1}^7{z}^{3m}\cdot \frac{1-z^{66m}}{1-z^{3m}}\\ & =22+\sum _{m=1}^7\frac{z^{3m}-z^{69m}}{1-z^{3m}}\\ & =22+\sum _{m=1}^7\frac{z^{3m}-1}{1-z^{3m}}\\ & =22+\sum _{m=1}^7(-1)\\ & =22-7=15.\end{array}$$Hence, $$\sum _{n=0}^{22}\frac{1}{1+z^n+z^{2n}}=\frac{1}{3}+15=\boxed{\frac{46}{3}}.$$