Is it possible to draw any regular polygon in such a way that its vertices be on the double grid, made of a rational and an irrational grid, superimposed, like in the image below?
If yes, what would be the step of the irrational grid?
(It is, evidently, possible to find such a double grid for drawing squares and equilateral triangles, as can be seen in the image below.)
Example of a double grid, rational and irrational (the step of the blue grid was taken $\sqrt 3$).
On
What I actually ended up doing below was merely to cover the case where certain pairs of vertices had to be at grid points of one of the grids. Allowing one vertex of the pair to be on one grid and one on the other opens up some possibilities; allowing the use of intersections of lines from different grids, along with some arbitrary offset of the grids' origins, opens up others. There might still be a way to make an argument like the following stick, perhaps using a polygon of more than ten sides. So I'll leave this here, but it is unfinished and possibly unproductive.
I interpret your question to be whether it is true for every $n \geq 3$ that there exists a regular polygon of $n$ sides that has all vertices on grid points.
Note that if $P,$ $Q,$ $R,$ and $S$ are grid points of a regular rectangular grid and the lines $PQ$ and $RS$ are parallel, then the ratio of distances $PQ:RS$ is rational. (Note that this result is true for any orientation of the grid that allows each line to intersect multiple grid points, not just the case where the lines are parallel to a line of the grid.)
Given a regular decagon $ABCDEFGHIJ$, consider the lines $AF,$ $BE,$ and $CD.$ These lines are all parallel. But the distances $BE$ and $CD$ are in the ratio $$\frac{BE}{CD} = \frac{\csc\left(3\pi/10\right)}{\csc\left(\pi/10\right)} = \frac12(3 - \sqrt5),$$ which is irrational, so the points $B$ and $E$ cannot be on the same grid as the points $C$ and $D.$
The distances $AF$ and $BE$ are in the ratio $$\frac{AF}{BE} = \frac{\csc\left(5\pi/10\right)}{\csc\left(3\pi/10\right)} = \frac14(1 + \sqrt5),$$ so $A$ and $F$ cannot be on the same grid as $B$ and $E.$
Finally, the distances $AF$ and $CD$ are in the ratio $$\frac{AF}{BE} = \frac{\csc\left(5\pi/10\right)}{\csc\left(\pi/10\right)} = \frac14(\sqrt5 - 1),$$ so $A$ and $F$ cannot be on the same grid as $C$ and $D.$
In short none of the three pairs of points can be grid points of the same grid as either of the other two pairs. You need at least three grids in order to be able to place all the points of a regular decagon on regular rectangular grids.
So if the pairs of vertices must fall on a regular grid point of one grid or the other, the answer is no, there is no mere pair of two grids that can provide vertices for regular polygons of all numbers of sides.
I suppose with rational and irrational grid you mean that there are a rational number $q$ and an irrational number $i$ such that for every point of the poligon, one of its coordinates is an integer multiple of either $q$ or $i$.
In this case, the answer is negative. Pick for example a square which has opposite vertices $(\sqrt{2},\sqrt{2}), (\sqrt{3},\sqrt{3})$.
Then, if this square followed what you say, $\sqrt{2}$ and $\sqrt{3}$ should be integer multiples of some irrational $i$, that is, $\sqrt{2}=ni$ and $\sqrt{3}=mi$, for some irrational $i$. However, this is impossible, because then we would have $\sqrt{3}=\frac{m}{n}\sqrt{2}$ for some integers $m,n$, which is false.