A 23x23 square 2.0

You can fill in an $11 \times 11$ block like so:

And you can make a $6 \times n$ rectangle for any $n \ge 2$ , by combining rows of $2 \times 2$ blocks, and one row of $3 \times 3$ blocks if $n$ is odd. This means that you can place the $11 \times 11$ block in one of 9 positions arranged in a $3 \times 3$ pattern, because each side of the block will have a multiple of 6 blocks remaining to fill in from that side to the very edge.

Consider the following colouring of the grid:

In this colouring, there are 92 of each group (A, B, C, D, E), except F, which has one less column, and has 69 squares instead.

Looking at the triplet $\Bigl((F-C) \bmod 6, (E-B) \bmod 6, (D-A) \bmod 6 \Bigr)$ , the initial value is $(1, 0, 0)$ . Placing a $3 \times 3$ block anywhere will change all the items in the triplet by 3, if it is an even number from $(0, 2, 4)$ it will be changed into an odd number from $(3, 5, 1)$ , and vice versa, while placing a $2 \times 2$ block will preserve odd values and even values. Since we start out with one odd value, the $1 \times 1$ block needs to be placed such that it makes that value even. this leaves two places for the $1 \times 1$ block, the F and C columns. If we place the $1 \times 1$ block in the C column, then we are left with the value $(2, 0, 0)$ , yet the sum of the first and last value minus the middle is always a multiple of 6 when you use $2 \times 2$ blocks, so it is unsolvable by using them, and using a $3 \times 3$ block won't help, as it will leave all the values odd.

We can apply this same colouring horizontally, and the intersection between the two colurings is the following green squares (labelled as B):