\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n.\left(n+1\right)}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..+\dfrac{1}{n}-\dfrac{1}{n-1}\)

\(A=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}=\dfrac{1999}{2000}\Rightarrow n.2000=\left(n+1\right).1999\)

\(\Leftrightarrow1999n+n=1999n+1999\Rightarrow n=1999\)