Let $\omega$ be a complex number such that
\[\omega + \frac{1}{\omega} = 1.\]Find all possible values of
\[\omega^n + \frac{1}{\omega^n},\]where $n$ is a positive integer.

Enter all possible values, separated by commas.
From the equation $\omega + \frac{1}{\omega} = 1,$ $\omega^2 + 1 = \omega,$ so
\[\omega^2 - \omega + 1 = 0.\]Then $(\omega + 1)(\omega^2 - \omega + 1) = 0,$ which expands as $\omega^3 + 1 = 0.$  Hence, $\omega^3 = -1.$

We divide into cases where $n$ is of the form $3k,$ $3k + 1,$ and $3k + 2.$

If $n = 3k,$ then
\[\omega^n + \frac{1}{\omega^n} = \omega^{3k} + \frac{1}{\omega^{3k}} = (\omega^3)^k + \frac{1}{(\omega^3)^k} = (-1)^k + \frac{1}{(-1)^k}.\]If $k$ is even, then this becomes 2, and if $k$ is odd, then this becomes $-2.$

If $n = 3k + 1,$ then
\begin{align*}
\omega^n + \frac{1}{\omega^n} &= \omega^{3k + 1} + \frac{1}{\omega^{3k + 1}} = (\omega^3)^k \omega + \frac{1}{(\omega^3)^k \omega} \\
&= (-1)^k \omega + \frac{1}{(-1)^k \omega} \\
&= (-1)^k \frac{\omega^2 + 1}{\omega} \\
&= (-1)^k \frac{-\omega}{\omega} \\
&= (-1)^k.
\end{align*}This can be $1$ or $-1$.

And if $n = 3k + 2,$ then
\begin{align*}
\omega^n + \frac{1}{\omega^n} &= \omega^{3k + 2} + \frac{1}{\omega^{3k + 2}} = (\omega^3)^k \omega^2 + \frac{1}{(\omega^3)^k \omega^2} \\
&= (-1)^k \omega^2 + \frac{1}{(-1)^k \omega^2} \\
&= (-1)^k \frac{\omega^4 + 1}{\omega^2} \\
&= (-1)^k \frac{-\omega + 1}{\omega^2} \\
&= (-1)^k \frac{-\omega^2}{\omega^2} \\
&= -(-1)^k.
\end{align*}This can be $1$ or $-1$.

Hence, the possible values of $\omega^n + \frac{1}{\omega^n}$ are $\boxed{-2,-1,1,2}.$