Let $n$ be a positive integer. Initially, a $2n \times 2n$ grid has $k$ black cells and the rest white cells. The following two operations are allowed:
1. If a $2 \times 2$ square has exactly three black cells, the fourth is changed to a black cell.
2. If there are exactly two black cells in a $2 \times 2$ square, the black cells are changed to white and white to black.

We aim to find the smallest positive integer $k$ such that for any configuration of the $2n \times 2n$ grid with $k$ black cells, all cells can be black after a finite number of operations.

The answer is $n^2 + n + 1$.

### Proof:
1. **Proof that $n^2 + n$ can fail:**
- Divide the board into $n^2$ $2 \times 2$ squares.
- For squares above the main diagonal of $2 \times 2$ squares, paint the upper right square black.
- For squares below the main diagonal, paint only the lower left square black.
- This configuration ensures that no $2 \times 2$ square has exactly three black cells, preventing the first operation from being applied. Thus, it is possible that not all cells can be turned black.

2. **Proof that $k \geq n^2 + n + 1$ works:**
- If we can somehow get the number of black cells to increase, we clearly win. The objective is to get two black cells to share an edge.
- We will prove a lemma: If $k > \frac{(m+1)(n-1)}{4}$ squares on an $m \times n$ grid are painted black, then we can get two squares to be adjacent.

Lemma Proof:
- Perform smoothing operations to rearrange black and white cells.
- Consider a diagonal of black cells and assign coordinates to them.
- Show that by smoothing operations, the assigned squares must be white.
- Calculate the number of black cells that are not on the leftmost row or bottom column.
- If $a \leq \frac{(m-1)(n-1)}{4}$, then the total number of black cells is at most $\frac{mn + m + n + 1}{4}$.
- Otherwise, show that the number of black squares on the leftmost column and bottom row is limited, leading to a contradiction if there are more than $\frac{(m+1)(n+1)}{4}$ black squares.

Final Blow:
- Induct on $m + n$.
- Show that if $k \leq \frac{(m+1)(n+1)}{4}$, then the number of black squares never increases.
- Use symmetry and inductive hypothesis to handle different cases.
- Conclude that $k \geq n^2 + n + 1$ ensures that all cells can be turned black.

Thus, the smallest positive integer $k$ such that for any configuration of the $2n \times 2n$ grid with $k$ black cells, all cells can be black after a finite number of operations is $n^2 + n + 1$.

The answer is: $\boxed{n^2 + n + 1}$.