7. Given a positive integer that has exactly 2017 positive divisors and is not coprime with 2018. What is the remainder when the sum of all such positive integers is divided by 2019?
7.2.

Notice,
$$
2018=2 \times 1009, 2019=3 \times 673 \text {, }
$$

and $673$, $1009$, and $2017$ are all prime numbers.
By the problem, a positive integer with exactly 2017 positive divisors must be the 2016th power of some prime number.
Thus, the sum to consider is $2^{2016} + 1009^{2016}$. At this point,
$$
\begin{array}{l}
2^{672} \equiv 1(\bmod 673) \\
\Rightarrow\left(2^{672}\right)^{3} \equiv 1(\bmod 673) \\
\Rightarrow 2^{2016} \equiv 1(\bmod 673), \\
1009^{672} \equiv 1(\bmod 673) \\
\Rightarrow\left(1009^{672}\right)^{3} \equiv 1(\bmod 673) \\
\Rightarrow 1009^{2016} \equiv 1(\bmod 673), \\
2^{2016} \equiv 1(\bmod 3), \\
1009^{2016} \equiv 1(\bmod 3) . \\
\text { Therefore, } 2^{2016} \equiv 1(\bmod 3 \times 673), \\
1009^{2016} \equiv 1(\bmod 3 \times 673) .
\end{array}
$$

Thus, the required remainder is 2.