Answer: A

Small Hint:

Both cubics have $n = -1$ as a root; factor each and cancel common factors

Big Hint:

The ratio simplifies to $\dfrac{n - 2}{n - 3} = 1 + \dfrac{1}{n - 3};$ it is an integer only when $n - 3$ divides $1$

Solution:

Factor both cubics: $f(n) = (n + 1)(n - 2)(n - 4)$ and $g(n) = (n + 1)(n - 3)(n - 4).$ Away from $n \in \{-1, 3, 4\},$ where $g$ vanishes or the ratio is $\tfrac{0}{0},$ the common factors cancel and $\dfrac{f(n)}{g(n)} = \dfrac{n - 2}{n - 3} = 1 + \dfrac{1}{n - 3}.$ That’s an integer only when $n - 3 = \pm 1,$ so $n = 2$ or $n = 4.$ But $n = 4$ kills $g,$ so only $n = 2$ survives, and the sum is $2.$ Therefore, the answer is A.