LCM(three numbers) = $1001$ and HCF = $1$.

Prime factorization: $1001 = 7 \times 11 \times 13$.

• For HCF = $1$, no prime ($7$, $11$, $13$) can divide all three numbers simultaneously.
• For LCM = $1001$, across the three numbers we must collectively include each of $7$, $11$, $13$ (at least once) with exponent $1$.

We can construct many valid unordered triples (positives, $1$ allowed), e.g.:

1. $(7, 11, 13)$
2. $(1, 7, 143)$ [$143 = 11 \times 13$]
3. $(1, 11, 91)$ [$91 = 7 \times 13$]
4. $(1, 13, 77)$ [$77 = 7 \times 11$]
5. $(7, 11, 143)$
6. $(7, 13, 77)$
7. $(11, 13, 91)$
8. $(1, 1, 1001)$
9. $(1, 1001, 1001)$

Each triple has HCF = $1$ and LCM = $1001$. Since we already have more than 8 distinct triples,
$n > 8$.

Therefore, the correct option is More than 8.