I am not sure whether 2 or more squares can cover a cell at one time,however I assume that it's possibe.After some tries on values of $n$,we point out that $n=2k+1,n=2k+2$ wants same amount of rectangles to cover.Let's go to the proof: Let's say $c_1,...c_n$ for columns and $r_1,r_2.....r_n$.Just a quick calculation follows that for even $n$ count of intersection cells are $\frac{n^2}{4}$ and for odd $n$ the count of deleted cells are $\frac{(n+1)^2}{4}$.And we also know that for even $n$ there are exactly $n$ completely empty columns and rows in sum and for odd n this number is $n-1$.Now,this is the case,take $n=2k$,so there are $2k$ empty columns and rows in sum.Then,when you take $n=2k+1$,you have $2k$ completely empty columns and rows in sum.Just take the rectangles $1Xn$ and $2k$ rectangles are enough.So,for even $n$ you need $n$ rectangles and you need $n-1$ rectangles for odd $n$.

Nuran2010 wrote: I am not sure whether 2 or more squares can cover a cell at one time,however I assume that it's possibe.After some tries on values of $n$,we point out that $n=2k+1,n=2k+2$ wants same amount of rectangles to cover.Let's go to the proof: Let's say $c_1,...c_n$ for columns and $r_1,r_2.....r_n$.Just a quick calculation follows that for even $n$ count of intersection cells are $\frac{n^2}{4}$ and for odd $n$ the count of deleted cells are $\frac{(n+1)^2}{4}$.And we also know that for even $n$ there are exactly $n$ completely empty columns and rows in sum and for odd n this number is $n-1$.Now,this is the case,take $n=2k$,so there are $2k$ empty columns and rows in sum.Then,when you take $n=2k+1$,you have $2k$ completely empty columns and rows in sum.Just take the rectangles $1Xn$ and $2k$ rectangles are enough.So,for even $n$ you need $n$ rectangles and you need $n-1$ rectangles for odd $n$. By the way,note that empty column or row means that no cell is removed.