The answer is affirmative. There exist $2018$ distinct positive numbers satisfying the required conditions:
$$a, 2a, \dots, 2018a, \text{ where } a = \left( \frac{2018 \cdot 2019 \cdot 4037}{6} \right)^4.$$
Note that $a$ is an integer number, since $2019$ is a multiple of $3$ and $2018$ is a multiple of $2$.

The sum of the squares of these numbers is
$$a^2 + (2a)^2 + \dots + (2018a)^2 = a^2(1^2 + 2^2 + \dots + 2018^2) = a^2 \left( \frac{2018 \cdot 2019 \cdot 4037}{6} \right) = \\ = \left( \frac{2018 \cdot 2019 \cdot 4037}{6} \right)^8 \left( \frac{2018 \cdot 2019 \cdot 4037}{6} \right) = \left( \left( \frac{2018 \cdot 2019 \cdot 4037}{6} \right)^3 \right)^3,$$
which is a perfect cube, and the sum of their cubes is
$$a^3 + (2a)^3 + \dots + (2018a)^3 = a^3(1^3 + 2^3 + \dots + 2018^3) = a^3 \left(\frac{2018 \cdot 2019}{2}\right)^2 = \\ = \left(\frac{2018 \cdot 2019 \cdot 4037}{6}\right)^{12} \left(\frac{2018 \cdot 2019}{2}\right)^2 = \left(\left(\frac{2018 \cdot 2019 \cdot 4037}{6}\right)^6 \left(\frac{2018 \cdot 2019}{2}\right)\right)^2,$$
which is a perfect square.