Let $ n=\prod_{i=1}^mp_i^{k_i}$, then $ \prod_i (k_i+1)=16$. Obviosly $ d_{8-k}d_{9+k}=n,k=0,1,..,7$. $ 18=d_6|n$, therefore $ p_1=2,p_2=3,k_2\ge 2, 2\le m\le 3.$ If $ m=2$, then $ n=2^33^3$ or $ n=2*3^7$ For first cfse $ d_6=8\not =18$. Second is solution. If $ m=3$, then $ n=2*3^3*p,p\ge 5$, $ d_1=1,d_2=2,d_3=3.$ If $ p<9$ $ d_6=9$. If $ p\ge 11$ $ d_4=6,d_5=9$ $ d_6=18\to p\ge 19$. We had divisors 27, 54, p,2p. If p<54, then must be $ d_9=54,d_8=54-17=37=p$, if p>54 $ d_8=54,d_9=54+17=71=p$. Therefore all solutions are $ n=2*3^7,n=2*3^3*37,n=2*3^3*71$.

Rust The first solution is false because d8=54 and d9=81 then the difference is d9-d8=27, a contradiction.

Yes. You are rite. I don't chek $ d_9-d_8=17$ for these case.

Slightly different finish: So as above, we get that $n=2 \cdot 3^3 \cdot p$ for some prime $p$. Then, we have that there exists $k$ with $k(k+17)=n=54p$ Considering modulo $27$ gives $k \equiv 0, 10 \pmod {27}$ In the former case, letting $k=27a$, we have $a(27a+17)=2p$. Since the two things are relatively prime, for there only to be prime factors of $p,2$ we must have $a=1,2$ (else there are at least 3 prime factors: a 2, and one further in each parenthetical part) $a=1$ yields $p=22$, which isn't prime, but $a=2$ yields $71$ which works. In the latter case, letting $k=27a+10$ yields $(a+1)(27a+10)=2p$. Again, we have $a+1 \leq 2 \implies a \leq 1$. But $a=1$ works, yielding $p=37$. Thus the only possible working values of $n$ are $\boxed{n=1998, 3834}$ (54 times 37 and 71 respectively). It can easily be verified that they work, so we're done.

@Mewto $k(k+17)=n=54p$ How did you know that $d_8,d_9$ don't have common prime divisors?