Solution:

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) $\overline{AB}$ if and only if a point on $\overline{AB}$ is a distance $r$ from the origin. $\overline{AB}$ 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 $\overline{AB}$ 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 and 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 $\overline{AB}$.