Note that $1001=7 \cdot 11 \cdot 13$, so the stated product must be a multiple of 7, as well as a multiple of 11, as well as a multiple of 13. There are 4 possible residues of $n$ modulo 11 for which the product is a multiple of 11; similarly, there are 4 possible residues of $n$ modulo 13 for which the product is a multiple of 13. However, there are only 3 possible residues of $n$ modulo 7 for which the product is a multiple of 7. Consider each of these $4 \cdot 4 \cdot 3=48$ possible triples of remainders. By the Chinese Remainder Theorem there is exactly one value of $n$ with $1 \leq n \leq 1001$ achieving those remainders, and exactly one value of $n$ with $16 \leq n \leq 1016$ achieving those remainders. Similarly, there is exactly one value of $n$ with $1017 \leq n \leq 2017$ with those same remainders. Hence there are 96 values of $n$ with $16 \leq n \leq 2017$ such that $(n-2)(n-0)(n-1)(n-7)$ is a multiple of 1001. It remains to check $n \in\{1,2,3, \ldots, 15\}$. Since the product must be a multiple of 7, we can narrow the set to $\{1,2,7,8,9,14\}$. The first 3 values work trivially, since the product is 0. It can be easily checked that none of the remaining values of $n$ yield a product which is a multiple of 11. Hence, the final answer is $96+3=99$.