\[ \frac{XB}{XA} = \frac{[OXB]}{[OXA]} = \frac{OB\times \sin \angle BOX}{OA\times \sin \angle AOX} \]Likewise \[ \frac{YC}{YD} = \frac{[OYC]}{[OYD]} = \frac{OC \times \sin \angle YOC}{OD \times \sin \angle YOD} \]Since $X, O, Y$ are collinear \[ \angle AOX = \angle COY\text{ and }\angle BOX = \angle DOY \]Therefore \begin{align*} \frac{XB}{XA} + \frac{YC}{YD} & = \frac{OB\times \sin \angle BOX}{OA\times \sin \angle AOX} + \frac{OC \times \sin \angle YOC}{OD \times \sin \angle YOD} \\ & = \frac{4}{3}\frac{\sin \angle BOX}{\sin \angle AOX} + \frac{5}{6}\frac{\sin \angle AOX}{\sin \angle BOX} \end{align*}Let $x = \frac{\sin \angle BOX}{\sin \angle AOX}$. Now \[ \frac{XB}{XA} + \frac{YC}{YD} = \frac{4}{3}x + \frac{5}{6}x^{-1} \]Using AM-GM we get \[ \frac{4}{3}x + \frac{5}{6}x^{-1} \geq 2 \times \sqrt{\frac{4}{3}x \times \frac{5}{6}x^{-1}} = \frac{2\sqrt{10}}{3} \]$\frac{4}{3}x + \frac{5}{6}x^{-1} = \frac{2\sqrt{10}}{3}$ iff \[ \frac{4}{3}x = \frac{5}{6}x^{-1} \implies x = \sqrt{\frac{5}{8}} \]Now we need to show that there exists angles $\angle AOX$ and $\angle BOX$ such that $x = \sqrt{\frac{5}{8}}$. This is easy to show. Set $\angle AOX = 90^{\circ}$ and $\angle BOX = \arcsin \sqrt{\frac{5}{8}} \approx 52.24^{\circ}$

Let $P$ is intersection point of line $OX$ and $(OAB)$, $Q$ is intersection point of line $OY$ with $(OCD)$. by angle-chasing, triangle $PAB$ similar triangle $QCD$. therefore, $$\dfrac{XB}{XA}+\dfrac{YC}{YD}=\dfrac{PB\cdot OB}{PA\cdot OA}+\dfrac{QC\cdot OC}{QD\cdot OD}=\dfrac{4PB}{3PA}+\dfrac{5QC}{6QD}=\dfrac{4PB}{3PA}+\dfrac{5PA}{6PB}\geq 2\sqrt{\dfrac{4}{3}\cdot\dfrac{5}{6}}=\dfrac{2\sqrt{10}}{3}$$Answer : $\textbf{33}$. (Figures that satisfy the conditions do in fact exist.)