Let $A_1A_2\ldots A_{2006}$ be a polygon such that $A_1A_{1004}$, $A_2A_{1005}$, $\ldots$, $A_{1003}A_{2006}$ and $M_1M_{1004}$, $M_2M_{1005}$, $\ldots$, $M_{1003}M_{2006}$ are concurrent at $O$, where $M_1, M_2, \ldots, M_{2006}$ are the midpoints of $A_1A_2, A_2A_3, \ldots, A_{2006}A_1$, respectively.

Note that $A_1A_{1004}$, $A_2A_{1005}$, $M_1M_{1004}$ are concurrent, so $A_1A_2 \parallel A_{1004}A_{1005}$. Similarly, we obtain that any opposite sides are parallel.

Now we have by applying the ratio theorem repetitively, we obtain
$$\frac{OA_1}{OA_{1004}} = \frac{OA_2}{OA_{1005}} = \dots = \frac{OA_{1003}}{OA_{2006}} = \frac{OA_{1004}}{OA_1} = \frac{OA_{1005}}{OA_2} = \dots = \frac{OA_{2006}}{OA_{1003}}$$
This implies that $O$ is the midpoint of every main diagonal. Therefore, all opposite sides are congruent and parallel.