Find the sum of those numbers

If $x$ is relatively prime to $2019$ , $2019-x$ will also be relatively prime to $2019$ . There are $\phi(2019)=1344$ numbers relatively prime to $2019$ , so there are $\frac{1344}{2}=672$ pairs of $201$ 9. So the answer is $2019 \times 672=1356768$ .

$\sum_{(n,2019)=1}^{1 \leq n \leq 2019} n = \sum_{n=1}^{2018}n - (3 \times 1+3 \times 2+\dots + 3 \times 672) - (673 \times 1+ 673 \times 2) =$

$\sum_{n=1}^{2018}n - 3(\sum_{n=1}^{672}n) - 673-2\times 673=\frac{2018\times 2019}{2}-3\frac{672 \times 673}{2}-673-1346 = 1356768$