Intersecting 17-sided polygons

First, note that each side of $A$ can intersect with no more that $16$ sides of $B$ . Indeed, consider the line that contains it. It divides the plane into two parts. If $B_iB_{i+1}$ intersects it for all $i$ from $1$ to $16$ , then $B_1$ and $B_{17}$ are in the same half-plane so $B_{17}B_1$ cannot intersect it. Therefore, the total number of intersections can be at most $16 \times 17 = 272.$

To construct an example with $272$ intersections, consider $34$ points uniformly distributed on the unit circle, with angles to the $x$ -axis being $\frac{2\pi k}{34,}$ for $k=0,1,...,33$ . We will call these points $P_k.$

The heptadecagons $A$ and $B$ will be "stars", with $A$ using even $k$ and $B$ using odd. Specifically, the edge from $A_i$ to $A_{i+1}$ or from $A_{17}$ to $A_1$ is the segment from $P_k$ to $P_{k + 16 \pmod{34}}$ for some $k$ . Then $B$ is obtained from $A$ by rotating at an angle of $\frac{2 \pi}{34}$ about the origin. The picture below illustrates this construction with pentagons instead of heptadecagons.

Note that all edges of $A$ and $B$ are chords that are "close" to a diameter. One can see that for a given edge of $A$ exactly one edge of $B$ does not intersect it: the one that is obtained from it by the symmetry around the origin. For example, for the chord $A_1A_2$ that connects points $P_0$ and $P_{16},$ at most one of the endpoints of an edge of $B$ can be among $P_1,P_3,...,P_{15}.$ If neither are there, then the only way to have a chord this close to the diameter using the points $P_{17},...,P_{33}$ is to use $P_{17}$ and $P_{33}.$

So, the final answer is $272$ .