\begin{array}{l}
\dfrac{n+1}{n-2} \in Z\\
\Rightarrow n+1 \vdots n-2\\
\Rightarrow n-2+3 \vdots n-2\\
\Rightarrow 3 \vdots n-2\\
\Rightarrow n-2 \in Ư(3)=\{1,-1,3,-3\}\\
\Rightarrow n \in \{3,1,-1,5\}\\
\text{Vậy với } n \in \{3,1,-1,5\} \text{ thì }\dfrac{n+1}{n-2} \in Z
\end{array}