[Solution] From $\frac{M N}{A B}+\frac{M N}{C D}$
$$
\begin{array}{l}
=\frac{M O+O N}{A B}+\frac{M O+O N}{C D} \\
=\frac{M O}{A B}+\frac{O N}{A B}+\frac{M O}{C D}+\frac{O N}{C D} \\
=\frac{D O}{D B}+\frac{C O}{A C}+\frac{A O}{A C}+\frac{O B}{D B} \\
=\frac{D O+O B}{D B}+\frac{C O+A O}{A C} \\
=2,
\end{array}
$$
$$
\therefore \quad \frac{1}{A B}+\frac{1}{C D}=\frac{2}{M N} \text {. }
$$

Therefore, the correct choice is (B).
18.115 In trapezoid $A B C D$, $A B / / C D, A C$ intersects $B D$ at $O$, $M O N / / A B$, and $M O N$ intersects $A D, B C$ at $M, N$. Then $\frac{1}{A B}+\frac{1}{C D}$ equals
(A) $\frac{1}{M N}$.
(B) $\frac{2}{M N}$.
(C) $\frac{3}{M N}$.
(D) $\frac{4}{M N}$.
(China Guangzhou, Wuhan, Fuzhou, and other five cities junior high school mathematics league, 1987)