Let $Q$ be a $(2n+1) \times (2n+1)$ board. The problem states that every $2 \times 2$ sub-board of $Q$ contains at most $2$ black cells. We need to determine the maximum number of black cells that can be accommodated on the board under this constraint.

To approach the solution, consider dividing the board into smaller sub-boards. Notice that each $2 \times 2$ sub-board must have black cells distributed in such a way that no sub-board has more than 2 black cells.

A useful strategy is to consider a checkerboard pattern where black cells are placed in one color of the checkerboard pattern, specifically, such that black cells do not share a $2 \times 2$ sub-board. By organizing black cells in a checkerboard fashion, prospective $2 \times 2$ boards will only contain black cells at opposite corners, thus having at most 2 black cells inside them.

Next, calculate the maximum number of black cells possible in such a configuration.

Since the board is of size $(2n+1) \times (2n+1)$, evaluate the number of positions:
- Consider every alternative position filled starting from the top-left corner, which we designate as $(1,1)$ being black. This results in filling every alternate row and alternate column solely with a single-color cell.

The pattern ensures that approximately half the squares are black and the other half are not colored (if filled correctly). Here's how to calculate the exact number of black cells:
- Count the number of black cells required per row blocking at most two columns. In the grid's odd-numbered size $(2n+1)$, the arrangement results in fulfilling $n+1$ rows been cluttered with mostly filled including the central line.
- Thus, there will be a total number of black cells calculated as:

$$(2n+1)(n+1)$$

This configuration maximizes the number of black cells while adhering to every $2 \times 2$ sub-grid containing no more than 2 black cells.

Thus, the maximum number of black cells that can fit this configuration is:
$$\boxed{(2n+1)(n+1)}$$