Cute problem Assume that there exists a column with $n$ grids,such that each grid is covered by exactly one square. Since these grids are covered by a $2*2$ square,they get covered in pairs,which is a contradiction as $n$ is odd. Thus each column contains at most $n-1$ such grids and so,the number of such grids on the board $\le n(n-1)$. Now,we show that the desired can be achieved. It can be easily achieved for a $3*3$ square.Now,consider an L shaped block,composed of two rectangles of length $n+2$ and width $2$ such that combining it with a $n*n$ block results in a $(n+2)*(n+2)$ block. We cover the grids in the block by successively placing the squares from left to right and up to down,with only the two squares at the turn ,overlapping at exactly one point. Thus ,we cover $8* \frac{n+1}{2} -2$ which gives us $4n +2$ such covered grids and hence we are done by induction.