Let $\mathrm{f}\left(\mathrm{x}\right)-2{\mathrm{x}}^{\mathrm{n}}+\lambda ,\lambda \in \mathrm{R},\mathrm{n}\in \mathrm{N}\text{ and }\mathrm{f}\left(4\right)=133,\mathrm{f}\left(5\right)=255$ . 255. Then the sum of all the positive integer
divisors of $\left(\mathrm{f}\right(3)-\mathrm{f}(2\left)\right)$ is

$\mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^{\mathrm{n}}+\mathrm{\lambda }\Rightarrow \mathrm{f}\left(4\right)=133,\mathrm{f}\left(5\right)=255$

$2(4{)}^{\mathrm{n}}+\mathrm{\lambda }=133,2(5{)}^{\mathrm{n}}+\mathrm{\lambda }=255$

$2(5{)}^{\mathrm{n}}-2(4{)}^{\mathrm{n}}=122$

$5^{\mathrm{n}}-4^{\mathrm{n}}=61\to \mathrm{n}=3$

$\begin{array}{l}\mathrm{f}\left(3\right)-\mathrm{f}\left(2\right)=2(3{)}^{\mathrm{n}}-2(2{)}^{\mathrm{n}}=2\left(27\right)-2\left(8\right)=54-16=38\end{array}$
Dividing 1, 2, 19, 38 sum =60