2019 is going to be a fine year for number theory!

Write $S=\sum_{k=1}^{\infty}\frac{1}{k^2}+\sum_{k=1}^{\infty}\frac{a}{(3k)^2}+\sum_{k=1}^{\infty}\frac{b}{(673k)^2}+\sum_{k=1}^{\infty}\frac{c}{(2019k)^2}$ , with $a,b,c$ to be determined.

If $\gcd(n,2019)=3$ then $1+a=3$ so $a=2$ . If $\gcd(n,2019)=673$ then $1+b=673$ so $b=672$ . If $\gcd(n,2019)=2019$ then $1+a+b+c=2019$ so $c=1344$ . Thus $S=\left(1+\frac{2}{3^2}+\frac{672}{673^2}+\frac{1344}{2019^2}\right)\zeta(2)=\left(1+\frac{2}{3^2}\right)\left(1+\frac{672}{673^2}\right)\zeta(2)=\frac{4989611}{2019^2}\zeta(2)$ . The answer is $\boxed{2019}$ .

his is the definition of Riemann zeta function.

$\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s} \ (*)$

here $s=2$

$gcd(n,2019)$ can be either $1,3,673,2019$

$\sum_{n=1}^{\infty}\frac{gcd(n,2019)}{n^2}=\sum_{gcd(n,2019)=1}\frac{1}{n^2}+\sum_{gcd(n,2019)=3}\frac{3}{n^2}+\sum_{gcd(n,2019)=673}\frac{673}{n^2}+\sum_{gcd(n,2019)=2019}\frac{2019}{n^2}=$

$\sum_{gcd(n,2019)=1}\frac{1}{n^2}+\sum_{gcd(k,673)=1}\frac{3}{(3k)^2}+\sum_{gcd(n,3)=1}\frac{673}{(673k)^2}+\sum_{k}\frac{2019}{(2019k)^2}=$

$\zeta(2)+\sum_{gcd(k,673)=1}\frac{3-1}{(3k)^2}+\sum_{gcd(n,3)=1}\frac{673-1}{(673k)^2}+\sum_{k}\frac{2019+1}{(2019k)^2}=$

$\zeta(2)+\frac{3-1}{3^2}\zeta(2)+\frac{673-1}{673^2}\zeta(2)+(\sum_{k}\frac{2019+1}{(2019k)^2}-\sum_{k}\frac{3-1}{(2019k)^2}-\sum_{k}\frac{673-1}{(2019k)^2})=$

$\zeta(2)+\frac{3-1}{3^2}\zeta(2)+\frac{673-1}{673^2}\zeta(2)+\frac{2019-3-673+3}{2019^2}\zeta(2)$

after summing up, one realises that the fraction is not reducible and, therefore, the denominator is $2019^2$ .

@Otto Bretscher , nice problems as always.