Given function

$f(x)=2x^n+\lambda,\quad \lambda\in \mathbb{R},\; n\in \mathbb{N}$

and

$f(4)=133,\qquad f(5)=255$

Use the given values

From $f(4)=133$: $2\cdot 4^n+\lambda=133 \quad \Rightarrow \quad 2\cdot 4^n+\lambda=133$

From $f(5)=255$: $2\cdot 5^n+\lambda=255$

Eliminate $\lambda$

Subtract the first equation from the second: $2\cdot 5^n-2\cdot 4^n=255-133=122$

$2(5^n-4^n)=122$

$5^n-4^n=61$

Find $n$

Since $61$ is small, test natural numbers:

• For $n=1$: $5^1-4^1=5-4=1$
• For $n=2$: $5^2-4^2=25-16=9$
• For $n=3$: $5^3-4^3=125-64=61$

Hence, $n=3$

Find $\lambda$

Using $f(4)=133$: $2\cdot 4^3+\lambda=133$ $2\cdot 64+\lambda=133$ $128+\lambda=133$ $\lambda=5$

So, $f(x)=2x^3+5$

Compute $f(3)-f(2)$

$f(3)=2\cdot 3^3+5=2\cdot 27+5=59$ $f(2)=2\cdot 2^3+5=2\cdot 8+5=21$

Therefore, $f(3)-f(2)=59-21=38$

Find the sum of positive divisors of $38$

Prime factorization: $38=2\times 19$

Positive divisors are: $1,\;2,\;19,\;38$

Their sum is: $1+2+19+38=60$

Conclusion

The required sum is $\boxed{60}$

So the correct option is A.