Lời giải:
$2n+3\vdots 3n+2$
$\Rightarrow 3(2n+3)\vdots 3n+2$
$\Rightarrow 6n+9\vdots 3n+2$
$\Rightarrow 2(3n+2)+5\vdots 3n+2$
$\Rightarrow 5\vdots 3n+2$
$\Rightarrow 3n+2\in \left\{1; -1; 5; -5\right\}$
$\Rightarrow n\in \left\{\frac{-1}{3}; -1; 1; \frac{-7}{3}\right\}$
Do $n$ nguyên nên $n\in \left\{-1;1\right\}$