Note that for each $d_i$ there exists a unique $d_j$ such that $d_id_j=n$, and holds $i+j=k+1$. So we have $k+1=19+20=39 \implies k=38$. Let $n=\prod_{i}^{k}p_i^{a_i}$ wich has exactly $\prod_{i=1}^{k}(a_i+1)=38$ divisors, therefore $a_i=1$, $a_2=18$ since $n$ has at least $2$ prime divisors. Now it is easy to see that $n=3\cdot 673^{18}$ and $n=3^{18} \cdot 673$.

$2019= 673 \cdot 3$. Now, $n=d_1 \cdot d_k = d_2 \cdot d_{k-1} \cdots = d_{19} \cdot d_{20} \implies 19+20=k+1$, so $k=38$. So $\tau(n)=38$. This means that $n=p^{37}$ or $n=pq^{18}$. Since $n=673^a \cdot 3^b \cdot x$, it implies that $n$ has at least $2$ prime divisors. Therefore, $n=pq^{18}$. Since the prime divisors of $n$ are $3,673$, $n=3 \cdot 673^{18}$ or $n=673 \cdot 3^{18}$.

enthusiast101 wrote: $2019= 673 \cdot 3$. Now, $n=d_1 \cdot d_k = d_2 \cdot d_{k-1} \cdots = d_{19} \cdot d_{20} \implies 19+20=k+1$, so $k=38$. So $\tau(n)=38$. This means that $n=p^{37}$ or $n=pq^{18}$. Since $n=673^a \cdot 3^b \cdot x$, it implies that $n$ has at least $2$ prime divisors. Therefore, $n=pq^{18}$. Since the prime divisors of $n$ are $3,673$, $n=3 \cdot 673^{18}$ or $n=673 \cdot 3^{18}$. Hey, it's same as @2above

n = d19 x d20 we know n = d(a)d(k+1-a) so k = 38 now since 2019 divides n <=> n has factors 3 and 673 Also we know (p+1)(q+1).... = 38 = 2 x 19, where p+1 , q+1 > 1 => (p+1)(q+1) = 2 x 19, where p and q are v_3(n) and v_673(n) Now possibilities p+1 = 2 , q+1 = 19; p+1 = 19, q+1 = 2 so answers ->673^18 x 3, 3^18 x 673..... Easily verifiable

Same as all above. $n$ has 38 divisors, now prime factorise and look at the answers.

I've seen this question multiple times in my math book (Always solved it by using that n has 38 divisors) and now I finally know where it is from!