Consider an ellipse and a set of lines whose number is odd. Number them 1, 2, ... , k (k odd).
Start from an arbitrary point G of the ellipse and construct a polygonal line with 2k+1 vertices.
G' --- >1' --- >2'--- >3'--- > .... --- >k'--- >1''--- >2''--- > ... --- >k''.
Where (1') is the other intersection point of the ellipse with the parallel from G to line (1). Then (2') is the other intersection point of the ellipse with the parallel from (1') to line (2), (3') is the other intersection point of the ellipse with the parallel from (2') to line (3) etc. By going two times through all lines you create a polygonal line that comes back to the starting point G.

[0_0] [0_1] [0_2]
[1_0] [1_1] [1_2]


Looks complicated but the proof is easy. Take a homography, transforming the ellipse to a circle and preserving the line at infinity. Such homographies respect parallels and coincidence of points, hence the whole problem transforms to a similar one for a circle.

The simpler case of circles is considered in the file ClosedPolygonalLines_Circles.html .

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