Four operations

The expression is equal to $\sqrt{2019^2+2\cdot 2019+1}=\sqrt{(2019+1)^2}=2020.$

$\text{FullSimplify}\left[\sqrt{(n-n)+n\,n+\frac{n}{n}+(n+n)}\right]\Rightarrow \sqrt{(n+1)^2}\Rightarrow \pm\left|\left|{n+1}\right|\right|$

$\pm 2020$

Take n as 2019, and then simplify. That will become:

$\sqrt { (n+n)+(n-n)+(n\times n)+(n/n) } =\sqrt { (2n)(n^{ 2 }) } factorize =\sqrt { (1+n)^{ 2 } } =\pm n+1 =\boxed{\pm 2020}$