Để n+3/n-3 là số tự nhiên thì
\(\left\{{}\begin{matrix}\dfrac{n+3}{n-3}>0\\n+3⋮n-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n\in\left(-\infty;-3\right)\cup\left(3;+\infty\right)\\n-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\end{matrix}\right.\)
\(\Leftrightarrow n\in\left\{4;5;6;9\right\}\)