14. In $\triangle A B C$, it is known that $A B=\sqrt{30}, A C=$ $\sqrt{6}, B C=\sqrt{15}$, point $D$ makes $A D$ bisect $B C$, and $\angle A D B=90^{\circ}$, the ratio $\frac{S_{\triangle A D B}}{S_{\triangle A B C}}$ can be written in the form $\frac{m}{n}$, where $m, n$ are coprime positive integers. Then $m+n=$
14.65.

Let the midpoint of $BC$ be $E$, and $AD=\frac{x}{2}$. By the median formula, we get $AE=\frac{\sqrt{57}}{2}$.
Thus, $(\sqrt{30})^{2}-\left(\frac{x}{2}\right)^{2}$
$$
=\left(\frac{\sqrt{15}}{2}\right)^{2}-\left(\frac{x}{2}-\frac{\sqrt{57}}{2}\right)^{2} \text {. }
$$

Solving for $x$ gives $x=\frac{81}{\sqrt{57}}$.
Therefore, $\frac{m}{n}=\frac{S_{\triangle A D B}}{2 S_{\triangle E B}}=\frac{AD}{2 AE}=\frac{27}{38}$.
Thus, $m+n=27+38=65$.