1. We are given four distinct primes $p_1, p_2, p_3, p_4$ such that $p_1 < p_2 < p_3 < p_4$ and $n = p_1 p_2 p_3 p_4$. We need to find all $n < 2001$ such that $d_9 - d_8 = 22$, where $1 = d_1 < d_2 < \cdots < d_{16} = n$ are the divisors of $n$.

2. Since $d_i \cdot d_{17-i} = n$ for all $1 \leq i \leq 16$, choosing $i = 8$ gives us $d_8 \cdot d_9 = n$.

3. Assume $q \in \{2, 11\}$ and $q \mid n$. Then $q \mid d_8 d_9$ implies $q \mid d_8$ and $q \mid d_9$, leading to $q^2 \mid n$. However, since $n$ is square-free, this is impossible. Therefore, $p_1 \neq 2$ and $p_i \neq 11$ for $i \in \{2, 3, 4\}$.

4. Assume $p_1 > 3$. Then $n = p_1 p_2 p_3 p_4 \leq 5 \cdot 7 \cdot 13 \cdot 17 = 7735 > 2000$, which is a contradiction. Hence, $p_1 \leq 3$. Since $p_2 \neq 2$, we have $p_1 = 3$.

5. Assume $p_2 > 5$. Then $p_4 = \frac{n}{p_1 p_2 p_3} \leq \frac{2000}{3 \cdot 7 \cdot 13} = \frac{2000}{273} < 8$, contradicting $p_4 > 11$. Hence, $p_2 \leq 5$, which means $p_2 = 5$.

6. Given $p_1 = 3$ and $p_2 = 5$, we have $p_3^2 < p_3 p_4 = \frac{n}{3 \cdot 5} = \frac{n}{15} \leq \frac{2000}{15} < 144 = 12^2$, yielding $p_3 < 12$. Therefore, $p_3 = 7$.

7. Thus, $n = 3 \cdot 5 \cdot 7 \cdot p_4 = 105 p_4$. This implies $105 p_4 \leq 2000$, so $p_4 \leq \frac{2000}{105} < 20$. Since $p_3 = 7 < p_4 \leq 19$ and $p_4 \neq 11$, we have $p_4 \in \{13, 17, 19\}$.

8. Since $d_8 d_9 = n$ and $n$ is odd, $d_8$ must be odd. Given $d_9 - d_8 = 22$, we have $d_8 d_9 - d_8^2 = 22 d_8$, leading to $d_8^2 + 22 d_8 = n$. Thus, $(d_8 + 11)^2 = 105 p_4 + 121$.

9. Since $d_8$ is odd, $4 \mid 105 p_4 + 121$, implying $4 \mid p_4 + 1$. Given $p_4 \in \{13, 17, 19\}$, we find $p_4 = 19$.

10. Therefore, $n = 105 \cdot 19 = 1995$. This means $d_8 = 5 \cdot 7 = 35$ and $d_9 = 3 \cdot 19 = 57$, yielding $d_9 - d_8 = 57 - 35 = 22$.

The final answer is $\boxed{ n = 1995 }$.