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Let $z=r{e}^{i\theta }$, where $r>0$ and $\theta \in \mathbb{R}$. Then $\bar{z}=r{e}^{-i\theta }$.

We have: $$\frac{\bar{z}}{z}+\frac{z}{\bar{z}}=\frac{r{e}^{-i\theta }}{r{e}^{i\theta }}+\frac{r{e}^{i\theta }}{r{e}^{-i\theta }}=e^{-2i\theta }+e^{2i\theta }=2\cos (2\theta )$$

We are told that $2\cos (2\theta )$ is a positive integer. The possible values for $2\cos (2\theta )$ are $1$ and $2$ (since $2\cos (2\theta )\le 2$ and must be a positive integer).

Case 1: $2\cos (2\theta )=2$

Then $\cos (2\theta )=1⟹2\theta =2\pi k$ for some integer $k$, so $\theta =\pi k$.

Thus, $z=r{e}^{i\pi k}=r(-1{)}^k$. So $z$ is a nonzero real number.

Case 2: $2\cos (2\theta )=1$

Then $\cos (2\theta )=\frac{1}{2}⟹2\theta =\pm \frac{\pi }{3}+2\pi k$ for some integer $k$.

So $\theta =\pm \frac{\pi }{6}+\pi k$.

Thus, $z=r{e}^{i(\pm \frac{\pi }{6}+\pi k)}$ for $r>0$ and integer $k$.

Therefore, all complex numbers $z\ne 0$ such that $z$ is a nonzero real number, or $z$ has argument $\pm \frac{\pi }{6}$ or $\pm \frac{7\pi }{6}$ (modulo $2\pi$), i.e., $z=r{e}^{i\theta }$ where $\theta =\pi k$ or $\theta =\pm \frac{\pi }{6}+\pi k$ for integer $k$ and $r>0$.