Solution:

Let $R A = x$. If the diagonals intersect at $X$, and the segment is $P Q$ with $P$ on $K R$, then $\triangle P K X \sim \triangle R K A$ and $\triangle O K X \sim \triangle R A X$ (by equal angles), giving $R A / P X = A K / X K = 1 + A X / X K = 1 + A R / O K = (x+12)/12$, so $P X = 12x/(12+x)$. Similarly $X Q = 12x/(12+x)$ also, so $P Q = 24x/(12+x) = 24 - \frac{288}{12+x}$. This has to be an integer. $288 = 2^{5} 3^{2}$, so it has $(5+1)(3+1) = 18$ divisors. $12 + x$ must be one of these. We also exclude the 8 divisors that don't exceed 12, so our final answer is 10.