Yes, it is possible.
Let $A_i$ be the set of all positive integers of the form $7^m(7k + i)$, where $m,k$ are non-negative integers, $i = 1,2,3,4,5,6$. We now show that $A_1,A_2,A_3,A_4,A_5,A_6$ satisfy the conditions of the problem.

Now suppose $x,y,z \in A_i$ and $x+2y=5z$; write $x,y,z$ respectively as
$$x = 7^{m_1}(7k_1 + i), \quad y = 7^{m_2}(7k_2 + i), \quad z = 7^{m_3}(7k_3 + i), \quad m_j, k_j \in \mathbb{N} \cup \{0\},$$
and let $\alpha = \min\{m_1,m_2,m_3\}$.

Dividing both sides of the equation $x+2y=5z$ by $7^\alpha$, and then reducing modulo 7, we obtain the congruence
$$e_1i + 2e_2i = 5e_3i, \quad (1)$$
where $e_1,e_2,e_3$ equal 0 or 1. But equation (1) only has the solution $e_1 = e_2 = e_3 = 0$, which is a contradiction. Therefore, positive integer solutions $(x,y,z)$ of $x+2y=5z$ can never all fall into the same $A_i$.