\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{1999}{2000}\)

\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{1999}{2000}\)

\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{1999}{2000}\)

\(2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1999}{2000}\)

\(2\left(\dfrac{1}{2}-\dfrac{1}{n+1}\right)=\dfrac{1999}{2000}\)

\(\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{1999}{2000}:2\)

\(\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{1999}{4000}\)

\(\dfrac{1}{n+1}=\dfrac{1}{2}-\dfrac{1999}{4000}\)

\(\dfrac{1}{n+1}=\dfrac{1}{4000}\)

\(\Rightarrow n+1=4000\)

\(n=4000-1\)

\(n=3999\)

Vậy n=3999