THE Fourteenth IMAR MATHEMATICAL COMPETITION

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 $a_0>0>a_2>a_{4n-2}>a_4>a_{4n-4}>\cdots >a_{2n-2}>a_{2n+2}>a_{2n}$, and $a_{2n+1}<a_{2n+3}<\cdots <a_{4n-1}<0<a_1<a_3<\cdots <a_{2n-1}$. Setting $A_{2k}=0\times a_{2k}$ and $A_{2k+1}=a_{2k+1}\times 0$, $k=0,\dots ,2n-1$, the polygon $A_0{A}_1\dots A_{4n-1}$ 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 $A_0{A}_1$ and $A_{2n+k-1}{A}_{2n+k}$ (respectively, $A_{4n-1}{A}_0$ and $A_k{A}_{k+1}$), $k=1,\dots ,2n-1$, and no other side, since the remaining sides all lie in the other two quadrants.