The square of the radius of a nice circle is the sum of the square of two integers. The nice circle of radius $r$ intersects (the open segment) $\overline{A B}$ if and only if a point on $\overline{A B}$ is a distance $r$ from the origin. $\overline{A B}$ consists of the points $(20, t)$ where $t$ ranges over $(15,16)$. The distance from the origin is $\sqrt{20^{2}+t^{2}}=\sqrt{400+t^{2}}$. As $t$ ranges over $(15,16), \sqrt{400+t^{2}}$ ranges over $(\sqrt{625}, \sqrt{656})$, so the nice circle of radius $r$ intersects $\overline{A B}$ if and only if $625<r^{2}<656$. The possible values of $r^{2}$ are those in this range that are the sum of two perfect squares, and each such value corresponds to a unique nice circle. By Fermat's Christmas theorem, an integer is the sum of two squares if an only if in its prime factorization, each prime that is $3 \bmod 4$ appears with an even exponent (possibly 0. ) In addition, since squares are 0,1 , or $4 \bmod 8$, we can quickly eliminate integers that are 3,6 , or $7 \bmod 8$. Now I will list all the integers that aren't 3,6 , or $7 \bmod 8$ in the range and either supply the bad prime factor or write "nice" with the prime factorization. 626: nice $(2 \cdot 313)$ 628: nice \left(2^{2} \cdot 157\right) 629: nice $(17 \cdot 37)$ 632: 79 633: 3 634: nice $(2 \cdot 317)$ 636: 3 637: nice \left(7^{2} \cdot 13\right) 640: nice \left(2^{7} \cdot 5\right) 641: nice $(641)$ 642: 3 644: 7 645: 3 648: nice \left(2^{3} \cdot 3^{4}\right) 649: 11 650: nice \left(2 \cdot 5^{2} \cdot 13\right) 652: 163 653: nice (653). There are 10 nice circles that intersect $\overline{A B}$.