If the unclear question means "how many ways can you partition the 20x20 unit region into multiple subregions each of which is a rectangle integer unit dimensions, with widths and lengths greater than zero?" then you've got a different answer.
There is $1$ way to partition it into $400$ different 1x1 squares.
There is $1$ way to partition in into $1$ large 20x20 square.
There are $0$ ways to partition it using $399$ {1x1} squares, because using $399$ {1x1} squares leaves a {1x1} area to be filled, which can only be filled by a {1x1} square yielding $400$ {1x1} squares, which is already counted above.
Using $398$ {1x1} squares, there are $20\times 19 \times 2 = 760$ ways to partition it: with $398$ 1x1 squares and $1$ 2x1 rectangle. The 2x1 rectangle can be posed vertically at $20$ different $x$-positions by $19$ different $y$ positions, or it can be posed horizontally at $19$ different $x$-positions by $20$ different $y$ positions, yielding $20 \cdot 19 \cdot 2$.
You can do similar combinatorics for using $397$ unit squares = $397$ {1x1} + $1$ {3x1}. Placing the long rectangle vertically yields $20 \ times 18$ ways, horizontally also yields $18 \times 20$ ways, totalling $720$ ways.
It gets more fun at $396$ unit squares, because the four squares you've removed from the first part of the solution above can be drawn as just $1$ {2x2} square, or as $2$ {1x2} squares (both of which can take on vertical or horizontal orientations), or as $1$ {4x1} square drawn either vertically or horizontally. The total number of these combinations can be calculated in a similar manner.
And keep going on up to using $0$ (zero) {1x1} squares: the single $1$ {20x20} square fitting in the area.
If you mean how many differently dimensioned rectangles could be drawn, one at a time, into a region of 20x20, then let
the x-dimension can vary from 1 to 20
the y-dimension can vary from 1 to 20
yielding $20\times 20=400$ different size rectangles if orientation matters. Orientation matters if a vertical {1x20}-sized rectangle is considered as being different from the horizontal {20x1}-sized rectangle.
If orientation does not matter, then in this manner of counting each rectangle is counted twice as an $m \times n$ rectangle and as an $n \times m$ rectangle, so divide that in half resulting
in $200$ different dimensioned rectangles could be drawn, one at a time, into a region of 20x20, if orientation does not matter.
There is also Scott Carnahan's approach in this question, which yields the answer $27$ using triangle numbers. That should also be the maximum number of rectangles yielded by my approach above; however, my answer is in the form of a heuristic rather than in closed form.
So of course, the answer depends exactly on what you mean specifically by the question.
