If $n \ge 3$ is even, $n = 2k$, then such a polygon exists since every regular $2k$-gon satisfies the condition.

If $n = 3$ or $n = 5$, then such a polygon does not exist. Indeed, no two sides in a triangle are parallel. If every side of a pentagon would be parallel to some other side, we could find three parallel sides and two of these should be adjacent. This is not possible.

For $n = 7$ such a polygon exists as shown by the figure.

Let us prove by induction that for odd positive integers $n$, $n \ge 7$, a polygon with the required property exists. Assume that for some integer $k$ a $k$-gon with this property exists. Choose a vertex at which the inner angle is less than $180^\circ$. Now, cut away a small parallelogram as shown in the figure. We obtain a $(k+2)$-gon which has the required property. A few examples are shown in the figure below.


$n = 9$

$n = 11$

$n = 13$