Assume that such an $ n$ gon $ A_1A_2\dots A_n$ exists. Note that $ A_{n-1}A_n\parallel A_{n-1-k}A_k$ for all $ k$. Now if the perpendicular bisector of (axis of symmetry of the $ n$ gon with respect to) $ A_{n-1}A_n$ intersects one of the sides (i.e. when $ n$ is even) then for some $ k$ we must have $ n-1-k+1=k$ hence $ n=2k$. Hence the diagonals parallel to $ A_{n-1}A_n$ are $ A_{n-1-k}A_k$ for $ k=1,2,\dots,\frac n2-2$. Therefore the total number of diagonals parallel to the sides are $ \frac 12n\left(\frac n2-2\right)$, since we counted each of the diagonals twice. But this is equal to $ \frac{n(n-4)}{4}$, which can never equal half of $ \frac{n(n-3)}{2}$, the total number of diagonals. [asy][asy]size(200); draw(MP("_{n-1-k-1}", (-1,1), W)--MP("_{n-1-k}", (0,0), SW)--MP("_{n-k (=k)}", (1.41,0), SE)--MP("_{n-k+1}", (2.41,1), E));[/asy][/asy] On the other hand, if the perpendicular bisector of (axis of symmetry with respect to) $ A_{n-1}A_n$ passes through one of the vertices (i.e. when $ n$ is odd) then for some $ k$ we need to have $ n-1-k=k$ i.e. $ n=2k+1$. Then the number of good diagonals is $ n\left(\frac{n-1}{2}-1\right)$, which is $ \frac{n(n-3)}{2}$, the total number of diagonals. Hence each of the diagonals of an $ n$ gon, where $ n$ is odd, is parallel to one of the sides. Therefore we conclude that no such polygon exists.