4 Place $r$ chess pieces on an $m \times n (m>1, n>1)$ chessboard, with at most one piece per square. If the $r$ pieces have the following property $p$: each row and each column contains at least one piece. However, removing any one of the pieces will cause them to no longer have the above property $p$. Find the maximum value of $r$, denoted as $r(m, n)$.
4. $r(m, n)=m+n-2$. On the one hand, by construction, $r=m+n-2$ is possible (all cells except the first cell in the first row and the first column are placed with chess pieces). Below is the proof: if $r2$ and $n>2$, then since there are at least $m+n-1>m$ chess pieces on the board, there must be a row with at least two chess pieces. Without loss of generality, assume the first row has the most chess pieces, a total of $t$ pieces, denoted as $a_{11}, a_{12}, \cdots, a_{1 t}(t>1)$. Then, no other chess pieces can be in the columns of these pieces; for example, if there is another chess piece in the column of $a_{11}$, then $a_{11}$ can be removed. Thus, removing the first row and the first $t$ columns, we are left with an $(m-1) \times (n-t)$ chessboard, which contains at least $m+n-1-t=(m-1)+(n-t)$ chess pieces. Note that $(m-1)+(n-t)=k-t-1<k$. If $n-t \geqslant 2$, applying the induction hypothesis to the $(m-1) \times (n-t)$ chessboard, we know that there exists a removable chess piece on the $(m-1) \times (n-t)$ chessboard. If $n-t \leqslant 1$, then the chess pieces are distributed on the first row or the $n$-th column. It is also easy to conclude that there is a removable chess piece. The proposition is proved.