$n^2-2004n.$

Solution: We call a square a “row-good square” if its number is greater than at least $2004$ numbers of squares from the same row. Since the least $2004$ numbers in each row do not belong to “row-good squares”, hence there are $n-2004$ “row-good squares” in each row. A “good square” must be a “row-good square”. Therefore there are $n(n-2004)$ “good squares” at most on the chessboard. Consider the following way of filling the chessboard with numbers $1,2,\cdots ,n^2$. First, we label some squares with “*” such that for row $i$ ($i=1,2,\cdots ,n$), only those squares in column $i, i+1,\cdots , i+2003$ ($i+k-n$ when $i+k>n, k\in\{1,2,\cdots ,2003\}$) are labeled. Then there are exactly $2004$ squares labeled in each row. Meanwhile there are also $2004$ squares labeled in each column. Indeed, when $1\le i\le 2003$, in column $i$, the squares in row $1,2,\cdots ,i,n+i-2003, n+i-2002,\cdots , n$ are labeled; when $i\ge 2004$, in column $i$, the squares in row $i-2003,i-2002,\cdots , i$ are labeled. Hence there are $2004$ squares labeled in each column. Now we put $1,2,3,\cdots ,2004n$ into the labeled squares, and put the rest numbers into the unlabeled squares. Since numbers in unlabeled squares are greater than numbers in labeled squares. Hence the unlabeled squares are all “good squares” and there are exactly $n^2-2004n$ “good squares”.