XXVII Olimpiada Matemática Rioplatense

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