Solution:

The answer is (B). The difference of the squares of two numbers that give the same remainder when divided by $k$ is a multiple of $k$: indeed if the two numbers are $a = n k + r$ and $b = m k + r$,
$$a^2 - b^2 = (a - b)(a + b) = (n - m) k (n k + m k + 2 r)$$
(in our case we have $r = 1$ for all $k$ from 2 to 13).

As a consequence, the product written by the goblin on the blackboard is certainly divisible by $13! = 2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7 \cdot 11 \cdot 13$, which is a multiple of $12^{5}$; therefore the goblin will certainly be able to divide by 12 for at least 5 days in a row always finding an integer result. But if the goblin were to choose for example the pairs $(2k+1, k+1)$ for $k = 4, 7, 10, 13$ and the pairs $(k+1, 1)$ for the remaining $k$'s, the differences of the squares (respectively $3k^2 + 2k$ and $k^2 + 2k = k(k+2)$) would contain in all exactly 5 factors of 3, therefore their product would turn out to be divisible by 12 no more than 5 times. But then in that case on the sixth day the division would produce a non-integer number; therefore we can count on 5 days, but no more.