, 2023

Let the row with the maximum number of red squares (let's assume it's the first row) have $a$ red squares, and the column with the maximum number of red squares (let's assume it's the first column) have $b$ red squares. If $a > b$, consider the $a$ columns in which the $a$ red squares of the first row are located. These columns must have distinct numbers of red squares and can only be chosen from $1, 2, \dots, b$. This leads to a contradiction. Similarly, we can show that $a$ cannot be less than $b$. Therefore, we conclude that $a = b$. Moreover, the number of red squares in the columns where the red squares of the first row are located must be a permutation of $1, 2, \dots, a$, and the number of red squares in the rows where the red squares of the first column are located must also be a permutation of $1, 2, \dots, a$. In other words, for any $1 \le i \le a$, there exist rows with exactly $i$ red squares and columns with exactly $i$ red squares.

Next, we prove that there exists a coloring that satisfies condition (1) and has $x_i$ rows with exactly $i$ red squares and $y_i$ columns with exactly $i$ red squares ($x_i, y_i \in \mathbb{Z}_{>0}$, $i = 1, 2, \dots, a$), if and only if there exists a coloring of an $a \times a$ grid such that the $i$th row has exactly $ix_i$ red squares and the $i$th column has exactly $iy_i$ red squares ($i = 1, 2, \dots, a$). (*).

First, we prove the necessity. If the coloring of a $70 \times 70$ grid $P$ satisfies condition (1), we construct an $a \times a$ grid $Q$ with the following coloring: for any $i, j \in \{1, 2, \dots, a\}$, the cell in the $i$th row and $j$th column of $Q$ is colored red if and only if $P$ has a red cell in a row with exactly $i$ red cells and a column with exactly $j$ red cells (denoted as $T(i, j)$). Note that in $P$, the number of red cells in a row with exactly $i$ red cells is $ix_i$, and the numbers of red cells in their corresponding columns are distinct. Therefore, $Q$ satisfies $T(i, j)$ if and only if there are exactly $ix_i$ red cells in column $j$, which implies that the $i$th row of $Q$ has exactly $ix_i$ red cells. Similarly, the $i$th column of $Q$ has exactly $iy_i$ red cells. Thus, the necessity is proved.

Next, we prove the sufficiency. Assume that the $a \times a$ grid $Q$ has exactly $ix_i$ red cells in the $i$th row and exactly $iy_i$ red cells in the $i$th column. We construct a $70 \times 70$ grid $P$ as follows: we label all rows of $P$ as $R(i, x)$ ($1 \le x \le x_i$) and all columns of $P$ as $C(i, y)$ ($1 \le y \le y_i$). Then, we color the cells of $P$ as follows: if the cell in the $i$th row and $j$th column of $Q$ is red, we mark $(i, j)$ as a good pair. For each good pair $(i_0, j_0)$, we define $v(i_0, j_0)$ as the index of $j$ (in ascending order) such that $(i_0, j)$ is a good pair, and $u(i_0, j_0)$ as the index of $i$ (in ascending order) such that $(i, j_0)$ is a good pair. Then, in $P$, we color the intersection of row $R(i_0, \lfloor \frac{v(i_0, j_0)}{i_0} \rfloor)$ and column $C(j_0, \lfloor \frac{u(i_0, j_0)}{j_0} \rfloor)$ red. In this way, each good pair $(i_0, j_0)$ is associated with only one pair of row $R(i_0, x)$ and column $C(j_0, y)$, satisfying condition (1). Moreover, for any row $R(i_0, x)$, the red cells correspond to $j$ values that are the $(x-1)i_0 + 1$st, $(x-1)i_0 + 2$nd, ..., $xi_0$th smallest $j$ among all good pairs $(i_0, j_0)$. Thus, this row has exactly $i_0$ red cells. Similarly, each column $C(j_0, y)$ has exactly $j_0$ red cells. Therefore, the sufficiency is proved, and thus (*) holds.

Returning to the original problem, let's first construct an example for $m = 32$. Consider a $64 \times 64$ grid $Q$, where the cell in the $i$th row and $j$th column is colored red if and only if $i + j \le 66$. This coloring ensures that each row of $Q$ has a decreasing number of red cells from 64 to 2, and each column has the same pattern. Now, let
$$a = 64, \quad x_{32} = x_{16} = x_8 = x_4 = x_2 = x_1 = y_{32} = y_{16} = y_8 = y_4 = y_2 = y_1 = 2,$$
while all other $x_i$ and $y_i$ are set to 1. According to (*) and the above construction, we know that there exists a grid $P$ that satisfies condition (1) and has two rows with exactly 32 red cells. Thus, we have shown that $m = 32$ meets the requirements of the problem.

Finally, we prove that $m$ cannot be greater than 32. Note that when $x_i \ge 2$, we have $2m \le a \le 69$, which implies $m \le 34$. We only need to show that $m$ cannot equal 33 or 34.
In this case, we only need to consider $a \in \{66, 67, 68, 69\}$. For the grid $Q$ and any $r, s \le a$, the difference between the total number of red cells in any $r$ rows and the total number of red cells in any $s$ columns is no more than the total number of red cells in those $r$ rows and the remaining $a-s$ columns, which is no more than $r(a-s)$.

If $a = 69$, then we have $x_m = 2$ and all other $x_i$ can only be 1. From calculating twice, we find that only $y_m = 2$ and all other $y_i$ can only be 1. In this case, in the grid $Q$, the top 5 rows with the maximum number of red cells have at least $69+68+67+66+66$ red cells, while the leftmost 5 columns with the minimum number of red cells have $1+2+3+4+5$ red cells. However,
$$(69 + 68 + 67 + 66 + 66) - (1 + 2 + 3 + 4 + 5) > 5 \cdot (69 - 5),$$
which leads to a contradiction!

If $a = 68$, similarly, we have
$$(68 + 67 + 66 + 66 + 66) - (1 + 2 + 3 + 4) > 4 \cdot (68 - 4),$$
which means that at least one of $y_1, y_2, y_3, y_4$ is not less than 2.
Since
$$(68 + 67 + 66 + 66 + 65 + 64 + 63 + 62) - (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) - 8 \cdot (68 - 8) > 4,$$
we have $\sum_{i=1}^{8}(y_i - 1) \ge 2$. Furthermore, notice that
$$(68 + 67 + 66 + 66 + 65 + \dots + 54) - (1 + 2 + \dots + 16) - 16 \cdot (68 - 16) > 4 + 8,$$
which implies $\sum_{i=1}^{16}(y_i - 1) \ge 3$. However, this contradicts the fact that $\sum_{i=1}^{68}(y_i - 1) = 2$.

If $a = 67$ (in this case, $m = 33$), similarly, we have
$$(67 + 66 + 66 + 66) - (1 + 2 + 3) > 3 \cdot (67 - 3),$$
which implies that at least one of $y_1, y_2, y_3$ is not less than 2. Additionally, we have
$$(67 + 66 + 66 + 65 + 64 + 63) - (1 + 2 + 3 + 4 + 5 + 6) - 6 \cdot (67 - 6) > 3,$$
which means that $\sum_{i=1}^{6}(y_i - 1) \ge 2$. Furthermore, we have
$$(67 + 66 + 66 + 65 + \dots + 57) - (1 + 2 + \dots + 12) - 12 \cdot (67 - 12) > 3 + 6,$$
implying that $\sum_{i=1}^{12}(y_i - 1) \ge 3$. Finally, we have
$$(67 + 66 + 66 + 65 + \dots + 45) - (1 + 2 + \dots + 24) - 24 \cdot (67 - 24) > 3 + 6 + 12,$$
which shows that $\sum_{i=1}^{24}(y_i - 1) \ge 4$. However, this contradicts the fact that $\sum_{i=1}^{67}(y_i - 1) = 3$.

If $a = 66$ (in this case, $m = 33$), we have two values of $ix_i$ equal to 66, indicating that all $iy_i$ are not less than 2, i.e., $y_1 \ge 2$. Similarly, we have
$$(66 + 66 + 65) - (1 + 2) - 3 \cdot (66 - 2) > 1,$$
which implies $\sum_{i=1}^{2}(y_i - 1) \ge 2$. Furthermore, we have
$$(66 + 66 + 65 + 64 + 63) - (1 + 2 + 3 + 4) - 5 \cdot (66 - 4) > 1 + 2,$$
implying $\sum_{i=1}^{4}(y_i - 1) \ge 3$. Additionally, we have
$$(66 + 66 + 65 + \cdots + 59) - (1 + 2 + \cdots + 8) - 9 \cdot (66 - 8) > 1 + 2 + 4,$$
which shows $\sum_{i=1}^{8}(y_i - 1) \ge 4$. Moreover, we have
$$(66 + 66 + 65 + \cdots + 51) - (1 + 2 + \cdots + 16) - 17 \cdot (66 - 16) > 1 + 2 + 4 + 8,$$
implying $\sum_{i=1}^{16}(y_i - 1) \ge 5$. However, this contradicts the fact that $\sum_{i=1}^{66}(y_i - 1) = 4$.

In conclusion, the maximum value of $m$ that satisfies the conditions is 32. $\square$