Answer (B): Note that $60 = 10 \cdot (-1) \cdot (-6)$, and the sum of these factors is $3$. It remains to show that no positive sum can be less than $3$. Such a sum would have to consist of one positive integer and two negative integers with smaller absolute value. If the positive integer is greater than or equal to $10$, then the sum is greater than or equal to $3$. The possible sets of factors in this case are $\{10, -6, -1\}$, $\{10, -3, -2\}$, $\{12, -5, -1\}$, $\{15, -4, -1\}$, $\{15, -2, -2\}$, $\{20, -3, -1\}$, $\{30, -2, -1\}$, and $\{60, -1, -1\}$. None of these sets of factors has a sum less than $3$.

The only other possible choices for the positive integer are $5$ and $6$, and in neither case is a positive sum possible. Indeed, if the positive integer is $5$, then the only possible set of factors is $\{5, -3, -4\}$. If the positive integer is $6$, then the only possible set of factors is $\{6, -5, -2\}$. In both of these cases, the sum is not positive.