Group the summands into pairs: the first one together with the last one, the second one together with the second last one, etc. Adding the members of each pair gives us
$$\frac{1}{\left(\frac{i}{j}\right)^2 + 1} + \frac{1}{\left(\frac{j}{i}\right)^2 + 1} = \frac{\frac{i^2}{j^2} + 1 + \frac{j^2}{i^2} + 1}{\left(\frac{i^2}{j^2} + 1\right) \left(\frac{j^2}{i^2} + 1\right)} = \frac{\frac{i^2}{j^2} + \frac{j^2}{i^2} + 2}{\frac{i^2}{j^2} + \frac{j^2}{i^2} + 2} = 1.$$
As there are $2019$ pairs in total, the sum of all numbers in the pairs is $2019$. Since every summand occurs twice, the desired sum is $\frac{2019}{2}$.