\(\frac{4n-5}{2n-1}=\frac{2\left(2n-1\right)-3}{2n-1}=\frac{2\left(2n-1\right)}{2n-1}-\frac{3}{2n-1}=2-\frac{3}{2n-1}\in Z\)
\(\Rightarrow3⋮2n-1\Rightarrow2n-1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow2n\in\left\{2;0;4;-2\right\}\)
\(\Rightarrow n\in\left\{1;0;2\right\}\left(n\in N\right)\)