Group the observed numbers into sets:
$$A_0 = \{1\}, A_1 = \{2, 3\}, A_2 = \{4, 5, 6, 7\}, A_3 = \{8, 9, \dots, 15\}, A_4 = \{16, 17, \dots, 31\}, \\ A_5 = \{32, 33, \dots, 63\}, A_6 = \{64, 65, \dots, 127\}, A_7 = \{128, 129, \dots, 255\}.$$
The number of elements of $A_k$ is $2^k$, for $k = 0, 1, \dots, 7$.
Notice that, for each $n$, numbers $n$ and $2n$ belong to $A_k$ and $A_{k+1}$ respectively for some $k$. If all numbers from the sets $A_1, A_3, A_5$ and $A_7$ are drawn, the number of these numbers is $2 + 8 + 32 + 128 = 170$ and none of them is twice as big as any other.
Let us show that it is not possible to choose more than $170$ numbers that satisfy the condition. Let $a_k$ numbers from $A_k$ be drawn, for $k = 0, 1, 2, \dots, 7$.
Observe the sets $A_k$ and $A_{k+1}$. For every $m \in A_k$, $2m$ is in $A_{k+1}$. The number of these pairs $(m, 2m)$ is $2^k$. Clearly at most one number is drawn from every pair. Except those numbers, set $A_{k+1}$ has another $2^k$ odd numbers, so at most $2^k + 2^k = 2^{k+1}$ numbers are drawn from $A_k$ and $A_{k+1}$.
That means that $a_0 + a_1 \le 2^1, a_2 + a_3 \le 2^3, a_4 + a_5 \le 2^5, a_6 + a_7 \le 2^7$.
Adding up those inequalities we get $a_0 + a_1 + \dots + a_7 \le 170$. Thereby the maximum possible $N$ is $170$.