Geometrization of the positive integers I'll explain first how I thought of the problem .
I thought that assigning each positive integer a point in the plane and then making some geometrical+number theoretical conditions on them is a cool idea so I made up the problem .

Version 1.
  Is it possible to assign each positive integer $n>1$ a point in the plane $P_n$ (assign different points for different numbers ) and for each positive integer $k>1$ a line $d_k$ such that not all the points are collinear , not all the lines coincide and for each $k>1$ a positive integer the multiple of $k$ points (that is the $P_{nk}$ points with $n$ variable ) are all on the line $d_k$.

It seems that this version is straightforward :
Take two different lines $d_a$ and $d_b$ so $P_{ab}$ and $P_{2ab}$ are both on $d_a$ and $d_b$ so the lines coincide so this type of construction is not possible . 
This led me to modify the problem and this is what I made :

Version 2 . Assign to each positive integer $n>1$ a point in the plane $P_n$ (assign different points for different numbers ) and for each positive integer $k>1$ a line $d_k$ .Denote by $A_{n,x}$ the number points between $P_n, P_{2n} , \ldots ,P_{xn}$ that lie on the line $d_n$ . For what numbers $c$ with $0 \leq c \leq 1$ there exists such a configuration of points and lines such that not all the points are collinear , not all lines coincide and for all $n>1$ and $x \geq 1$ we have $A_{n,x} \geq cx$ (this is like a density ) .

Note that for $c=0$ there clearly is a configuration and for $c=1$ we've shown there aren't any so i think (intuition) there is some $c_0$ such that for $c\leq c_0$ there is a configuration and for $c>c_0$ there isn't .
This version may also be straightforward because there are lots of conditions for such a configuration but I haven't managed to find a proof .
What I did :
Noticed that if $cx>x-1$ that is $x<\frac{1}{1-c}$ we have necessarily $A_{n,x}=x$ for every such $x$ and every $n$ . Then I let $c>\frac{3}{4}$ so $\frac{1}{1-c}>4$ so every $x \leq 4$ has this property . Then through an induction argument I show that the lines $k$ and $k\cdot 2^i \cdot 3^j$ are the same for every $k$ , $i$ and $j$ . 
I guess this generalizes (though I haven't tried ) :If $c>\frac{n-1}{n}$ and $p_1,\ldots p_l$ are the primes at most $n$ then the lines $k$ and $kp_1^{i_1}\ldots p_l^{i_l}$ coincide for every possible exponents . This is practically equivalent with : If $a \leq n$ then the lines $d_k$ and $d_{ak}$ coincide for every k.
Maybe this helps but I haven't figured out how .
I'd appreciate any possible help .Thanks everyone .

First, it is clear that if the answer is positive for some $c_0$, then it must be for every $c < c_0$, too: if we keep the same configuration we have $A_{n,x} \geq c_0 x > cx$. Similarly, if the answer is negative for $c_0$ it must be negative for every $c > c_0$, too.
We will now prove that the answer is positive for every $c < 1$. Indeed, let $n$ be any prime number such that
$$
\frac{n - 1}{n} \geq c
$$
and consider two parallel lines: one for the multiples of $n$ and one for every other number.
Now, clearly for every $k \geq 1$ we have $A_{kn,x} = x > cx$. If instead $m$ is not a multiple of $n$ we have
$$
A_{m,x} = x - \left\lfloor \frac{x}{n} \right\rfloor \geq x - \frac{x}{n} = \left(\frac{n-1}{n}\right) x \geq cx.
$$

As a side note, you are quite right in thinking that this problem hides some notion of density. This answer hinges heavily on the fact that the natural density of $n\Bbb{N}$ in $\Bbb{N}$ is precisely $\frac{1}{n}$.

Update: Interestingly, this argument can be tweaked to construct a solution with countably many lines. Indeed, start by choosing a sequence of primes $\{p_i\}_{i \geq 1}$ such that
$$
1 - \sum_{i = 1}^{\infty} \frac{1}{p_i} > c.
$$
For example, we could simply choose $p_i > n^i$ for $n$ large enough, because this gives
$$
\sum_{i = 1}^{\infty} \frac{1}{p_i} < \sum_{i = 1}^{\infty} \frac{1}{n^i} = \frac{1}{n-1}.
$$
Now, for every $i \geq 1$ put every number divisible by $p_i$ but not $p_j$ for every $j \neq i$ on, say, the vertical line through $(p_i,0)$ in the Cartesian plane, and put every other number on the vertical line through $(1,0)$. Then for every positive integer $m$ we have
$$
\begin{align}
A_{m,x}
&\geq x - \sum_{i = 1}^{\infty} \left\lfloor \frac{x}{p_i} \right\rfloor\\
&\geq x - \sum_{i = 1}^{\infty} \frac{x}{p_i}\\
&= \left( 1 - \sum_{i = 1}^{\infty} \frac{1}{p_i} \right) x\\
&> cx.
\end{align}
$$
