Ta có :
$\dfrac{m}{n} = 1+\dfrac{1}{2}+\dfrac{1}{3}+..+\dfrac{1}{1998} $
$ = (1+\dfrac{1}{1998} ) + (\dfrac{1}{2}+\dfrac{1}{1007})+...+(\dfrac{1}{999} +\dfrac{1}{1000})$
$=\dfrac{1999}{1998}+\dfrac{1999}{2.1007}+...+\dfrac{1999}{999.1000} $
$=1999.(\dfrac{1}{1998}+\dfrac{1}{2.1007}+...+\dfrac{1}{999.1000}) \vdots 1999$
Nên $m \vdots 1999$