The answer is in the affirmative. To describe the configuration, fix a coordinate frame and let $a_0, a_1, \dots, a_{4n-1}$ be real numbers such that $a_0 > 0 > a_2 > a_{4n-2} > a_4 > a_{4n-4} > \dots > a_{2n-2} > a_{2n+2} > a_{2n}$, and $a_{2n+1} < a_{2n+3} < \dots < a_{4n-1} < 0 < a_1 < a_3 < \dots < a_{2n-1}$. Setting $A_{2k} = 0 \times a_{2k}$ and $A_{2k+1} = a_{2k+1} \times 0$, $k = 0, \dots, 2n-1$, the polygon $A_0A_1 \dots A_{4n-1}$ and the origin satisfy the condition in the statement: the x-axis (respectively, y-axis) contains all vertices of odd (respectively, even) rank and no other points on the boundary; and every line through the first and third (respectively, second and fourth) quadrants crosses the sides $A_0A_1$ and $A_{2n+k-1}A_{2n+k}$ (respectively, $A_{4n-1}A_0$ and $A_kA_{k+1}$), $k = 1, \dots, 2n-1$, and no other side, since the remaining sides all lie in the other two quadrants.