7.3 Given that a closed broken line is drawn along the grid lines of a square grid paper, consisting of 14 segments and with at most one edge of the broken line on each grid line, how many self-intersection points can this broken line have at most?

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[Solution] Since the polyline is drawn along the grid lines, its horizontal and vertical edges are alternately arranged, so among the 14 edges, there must be 7 horizontal edges and 7 vertical edges. Number the horizontal edges from top to bottom as 1 to 7. It is easy to see that there are no self-intersection points on the 1st horizontal edge, at most 2 self-intersection points on the 2nd horizontal edge, and at most 4 self-intersection points on the 3rd horizontal edge. Similarly, there are at most 0, 2, and 4 self-intersection points on the 7th, 6th, and 5th horizontal edges, respectively. For the 4th horizontal edge, there is one vertical edge connected to each end, so there are at most 5 self-intersection points. Therefore, the entire polyline has at most 17 self-intersection points.

On the other hand, the polyline in the right figure has 14 edges, each on a grid line, and has a total of 17 self-intersection points.
In summary, this polyline has at most 17 self-intersection points.