A beautiful problem from a Hungarian competition

We will prove that $36$ can only be in a corner. It's easy to see that every even number will be placed on the same colour. We have nine numbers from $1$ to $36$ which have the $4$ divisor: $4, 8, 12, 16, 20, 24, 28, 32, 36$ . We can suppose that even numbers will be placed on the red cells. In the figure we dismembered the table to nine pieces of $2\times 2$ square. There cannot be two-divisible by $4$ -numbers in one $2\times 2$ square, so in every $2\times 2$ square there will be exactly one number which is divisible by $4$ . The $6\times 6$ grid has a rotational symmetry, so we can suppose that in the middle $2\times 2$ square this number (which is divisble by $4$ ) is in the F cell. Then in the $A, B, C, E, G, I, J, K$ cells there must not be a number, which is a multiply of $4$ . So in the $IJMN$ $2\times 2$ square the number, which is divisible by $4$ , can only be in cell $N$ (because $M$ is a blue cell). Similarly in the $CDGH$ $2\times 2$ square the only possible place for the number, which is divisble by $4$ , is cell $H$ . Now we can see that in the $KLOP$ $2\times 2$ square the number, which is a multiply of $4$ , can't be in $K, L, O$ cells, so it must be on the $P$ cell. We will show that the only number, which can be on the $P$ cell, is $36$ . If it were different (for example $4*a$ ) then $1<4*a<37$ would be true, so $4a+2$ and $4a-2$ would have to be placed on the grid. These numbers can't be on the $N, H$ cells (because $4$ is not a divisor of $4a-2$ and $4a+2)$ , so in cell $K$ there would have to be $2$ numbers, which would not be possible. So $36$ has to be in a corner, and there are $4$ corners. A possible numbering:

This is a comment on Aron's very elegant solution. I came up with the correct solution using a related semi-empirical approach.

However, I noticed that Aron's solution had the same arrangement of 4-multiples as mine did across the entire board - i.e. 2 regimented columns of 3 on the left, a 3rd column offset at the right edge, separated by 2 columns with no multiples of 4. My actual numbers and orientation were different of course, but the pattern of 4-multiples was identical. I wonder whether this is the only valid arrangement. Anyone?