Example 3 In a $19 \times 89$ chessboard, what is the maximum number of chess pieces that can be placed so that no $2 \times 2$ rectangle contains more than 2 pieces?
Analysis and Solution Note that 19 and 89 are both odd numbers, so we can consider a general $(2m-1) \times (2n-1)$ chessboard (where $m$ and $n$ are at least one greater than 1). We can consider induction on $m$. Let the maximum number of pieces that can be placed on the chessboard be $r_m$.

(1) When $m=1$, each cell can have a piece, so $r_1 = 2n-1$.

Figure 13-4
(2) When $m=2$, by the placement shown in Figure 13-4, we can conjecture: $r_2 = 2(2n-1)$, i.e., a $3 \times (2n-1)$ chessboard can have at most $4n-2$ pieces.

Since a $3 \times (2n-1)$ chessboard needs to remove two rows to become a $1 \times (2n-1)$ chessboard. How many pieces can be in the two rows that are removed? For this, we need to study a $2 \times (2n-1)$ chessboard, which is a scenario we have skipped. This scenario may be helpful for studying the $3 \times (2n-1)$ chessboard. Therefore, we look back at the $2 \times (2n-1)$ chessboard.

From Figure 13-5, in a $2 \times (2n-1)$ chessboard, $r \leq 2n$, and the equality can only be achieved in the unique way shown in Figure 13-5.

In fact, the first column can have at most two pieces, and the subsequent $2n-2$ columns can be divided into $n-1$ $2 \times 2$ chessboards, each of which can have at most 2 pieces. Therefore, $r \leq 2 + 2(n-1) = 2n$. If $r = 2n$, then the first column must have two pieces, and each $2 \times 2$ chessboard must have exactly 2 pieces. Thus, the first column has two pieces, and the second column has no pieces, so the third column has two pieces. This continues, so all odd-numbered columns have two pieces, and all even-numbered columns have no pieces. The equality is achieved in a unique way.

Now consider a $3 \times (2n-1)$ chessboard. We need to prove that $r \leq 2(2n-1)$. Naturally, we should divide the $3 \times (2n-1)$ chessboard into a $2 \times (2n-1)$ chessboard (referred to as $A$ board) and a $1 \times (2n-1)$ chessboard (referred to as $B$ board).

From the previous discussion, we know that $r_A \leq 2n$ and $r_B \leq 2n-1$. Therefore,
$$r = r_A + r_B \leq 4n-1$$

If (1) holds with equality, then $r_A = 2n$ and $r_B = 2n-1$. In this case, the placement of pieces on the chessboard is as shown in Figure 13-6, but there is a $2 \times 2$ square with 3 pieces. This is a contradiction. Therefore, $r \leq 4n-2$.

There is a flaw in the above proof: to have a $2 \times 2$ square, $n > 1$ must hold, i.e., the chessboard must have at least two columns. Therefore, we need to optimize the assumption: $m \leq n$. In this case, $n \geq m > 1$ must hold.

In general, for a $(2m-1) \times (2n-1)$ chessboard, if $m \leq n$, we prove that $r \leq m(2n-1)$.

We use induction on $m$. When $m=1$, the conclusion is obviously true. Assume the conclusion holds for natural numbers less than $m$, and consider a $(2m-1) \times (2n-1)$ chessboard. We divide it into an $A$ board: a $2 \times (2n-1)$ chessboard and a $B$ board: a $(2m-3) \times (2n-1)$ chessboard. From the previous discussion and the induction hypothesis, we know that $r_A \leq 2n$ and $r_B \leq (m-1)(2n-1)$, so
$$r = r_A + r_B \leq 2n + (m-1)(2n-1) = m(2n-1) + 1$$

If (2) holds with equality, then $r_A = 2n$ and $r_B = (m-1)(2n-1)$. From $r_A = 2n$, we know that the first row of the entire $(2m-1) \times (2n-1)$ chessboard has exactly $n$ pieces. Thus, the remaining $2m-2$ rows can be divided into $m-1$ $2 \times (2n-1)$ chessboards, each of which has no more than $2n$ pieces. Therefore, the number of pieces on the chessboard $r \leq n + 2n(m-1) = 2mn - n \leq 2mn - m = m(2n-1)$, which contradicts $r = m(2n-1) + 1$. Therefore, (2) does not hold with equality. That is, $r \leq m(2n-1)$. The proposition is proved. $\square$

Finally, place a piece in each cell of the odd-numbered rows of the chessboard, and we have $r = m(2n-1)$, so $r_m = m(2n-1)$. In particular, let $m=10$ and $n=45$, then the $19 \times 89$ chessboard can have at most 890 pieces.