If we let $p_1 < p_2 < p_3 < p_4$, we have that $d_9 = p_2p_4$ and $d_8 = p_2p_3$ or $p_1p_4$ If $d_8 = p_2p_3$, then we have that $p_2(p_4-p_3) = 22$ which means $p_2 = 11$. But this means $n = p_1p_2p_3p_4 > (2)(11)(11)(11) = 2662$, which is impossible. So, $d_8 = p_1p_4$ which means $p_4(p_2-p_1) = 22 \implies p_4 = 11$ and $p_2 - p_1 = 2$. So, $(p_1,p_2)$ can only be $(3,5)$ or $(5,7)$. In the first case, we have that $n = (3)(5)(11)(p_4) = 165p_4 \ge 165(13) = 2145$, which is impossible. In the second case, we have that $n > 5005$, which is again impossible. So, there are no such $n$

We have given four primes $(1) \;\; p_1<p_2<p_3<p_4$ and an integer $(2) \;\; n = p_1p_2p_3p_4$. The problem is to find all $n<2001$ s.t. $(3) \;\; d_9 - d_8 = 22$, where $(4) \;\; 1 = d_1 < d_2 < \cdots < d_{15} < d_{16} = n$ are the positive divisors of $n$. Solution: We know that $d_i \cdot d_{17-i} = n$ for all $1 \leq i \leq 16$. Hence by choosing $i=8$, we obtain $(5) \;\; d_8 \cdot d_9 = n$. Assume $q \in \{2,11\}$ and $q | n$. Then $q | d_8d_9$ by equation (5), which combined with equation (3) give us $q | d_8$ and $q | d_9$, yielding $q^2 | d_8d_9$, i.e. $q^2 | n$, which is impossible since $n$ is squarefree according to condition (1) and equation (2). Hence $p_1 \neq 2$ and $p_i \neq 11$ for $i \in \{2,3,4\}$. Assume $p_1>3$. Then by condition (1) and formula (2) $n = p_1p_2p_3p_4 \leq 5 \cdot 7 \cdot 13 \cdot 17 = 7735 > 2000$, a contradiction which implies $p_1 \leq 3$. Thus, since $p_2 \neq 2$, we have $p_1=3$. Next assume $p_2>5$. Then by equation (2) ${\textstyle p_4 = \frac{n}{p_1p_2p_3} \leq \frac{2000}{3 \cdot 7 \cdot 13} = \frac{2000}{273} < 8}$, contradicting $p_4>11$ by condition (1) where $p_1=3$. Hence $p_2 \leq 5$, which means $p_2=5$ (since $3 = p_1 < p_2 \leq 5$). Hence by condition (1) and equation (2) ${\textstyle p_3^2 < p_3p_4 = \frac{n}{p_1p_2} = \frac{n}{3 \cdot 5} = \frac{n}{15} \leq \frac{2000}{15} < 144 = 12^2}$, yielding $p_3<12$. Therefore, since $p_2=5 < p_3 < 12$ and $p_3 \neq 11$, we obtain $p_3=7$. Hence $n = 3 \cdot 5 \cdot 7 \cdot p_4 = 105p_4$ by quation (2). This means $105p_4 \leq 2000$, implying ${\textstyle p_4 \leq \frac{2000}{105} < 20}$, which (since $p_3=7 < p_4 \leq 19$ and $p_4 \neq 11$) yields $(6) \;\; p_4 \in \{13,17,19\}$. Observe that $d_8d_9=n$ implies (since $n$ is odd) $d_8$ is odd. Moreover by equation (3) $d_9 - d_8 = 22$ $d_8d_9 - d_8^2 = 22d_8$ $d_8^2 + 22d_8 = n$. $(d_8 + 11)^2 = 105p_4 + 121$ Consequently $4 \mid 105p_4 + 121$ (since $d_8$ is odd), yielding $4 \mid p_4 + 1$, which combined with condition (6) give us $p_4=19$. Hence $n = 105p_4 = 105 \cdot 19 = 1995$, which means $d_8 = 5 \cdot 7 = 35$ and $d_9 = 3 \cdot 19 = 57$, yielding $d_9 - d_8 = 57 - 35 = 22$. Conclusion: The only positive integer $n<2001$ satisfying conditions/equations (1)-(4) is $n=1995$.