Solution:

We proceed by induction on $N$.

Base case: $N = 1$. The single black square can be cut out as a $1 \times 1$ square. The black area is $1$, which is between $1/5$ and $4/5$ of $1$ (since $1 = 1$). So the base case holds.

Inductive step: Suppose the statement is true for all numbers less than $N$. Consider $N$ black squares.

Let $k$ be the largest integer such that $k^2 \leq N$. Consider all possible $k \times k$ squares on the grid. Each such square contains at most $k^2$ black squares. Among all $k \times k$ squares, choose one that contains the largest number $m$ of black squares. Then $m \geq N / (n - k + 1)^2$, where $n$ is the side length of the grid (but since the paper is large, we can always find such a square).

If $m \geq k^2 / 5$, then $m \leq k^2$ and $m \geq k^2 / 5$. If $m \leq 4k^2 / 5$, then we can cut out this $k \times k$ square, and the black area is between $1/5$ and $4/5$ of its total area. Remove these $m$ black squares, and apply the induction hypothesis to the remaining $N - m$ black squares.

If $m > 4k^2 / 5$, then $m$ is close to $k^2$. Cut out this $k \times k$ square. The black area is more than $4/5$ of its total area, but less than or equal to $k^2$. In this case, we can cut out a larger square. Consider a $(k+1) \times (k+1)$ square containing the $k \times k$ square. The number of black squares in this $(k+1) \times (k+1)$ square is at most $N$, and at least $m$. If the number of black squares in this $(k+1) \times (k+1)$ square is between $1/5$ and $4/5$ of its area, we are done. If not, repeat the process with larger squares.

Since the total number of black squares is $N$, and the area of the paper is much larger, we can always find such squares. At each step, we remove a square with the required property, and the process terminates after finitely many steps.

Therefore, it is always possible to cut out a set of disjoint square pieces of paper so that all the black squares are removed and the black area of each piece is between $1/5$ and $4/5$ of its total area.