Problem 1252

1. Lemma Proof:
We need to show that the $n$ distinct numbers of red squares in rows and columns must be either $\{1, 2, \ldots, n\}$ or $\{0, 1, 2, \ldots, n-1\}$.

- Suppose some $1 < i < n$ is not contained in the rows (or columns). This implies that both $0$ and $n$ must be contained in the rows (or columns).
- Therefore, the $n$ distinct numbers of red squares for each column (or row), denoted as $c(i)$ for $1 \leq i \leq n$, must satisfy $1 \leq c(i) \leq n-1$, which gives us $n-1$ distinct numbers in total.
- This is a contradiction because we need $n$ distinct numbers. Hence, the distinct numbers must be either $\{1, 2, \ldots, n\}$ or $\{0, 1, 2, \ldots, n-1\}$.

2. Equivalence of Cases:
- The number of ways to color the board using $\{1, 2, \ldots, n\}$ is equal to the number of ways using $\{0, 1, 2, \ldots, n-1\}$.
- This is because we can consider white squares instead of red squares, effectively swapping the roles of red and white.

3. Counting the Ways:
- We denote $(r_1, r_2, \ldots, r_n; c_1, c_2, \ldots, c_n)$ as a possible way of coloring, where $r_i$ denotes the number of red squares in the $i$-th row, and $c_i$ denotes the number of red squares in the $i$-th column.
- We need to prove that there exists a unique way satisfying $r_i = a_i$ and $c_i = b_i$ for any distinct $1 \leq a_i, b_i \leq n$.

4. Existence Proof:
- Consider the original way $(m_{ij})_{1 \leq i, j \leq n}$ where $m_{ij}$ is red if $i \leq j$.
- Rearrange the rows and columns so that the way satisfies $(a_1, a_2, \ldots, a_n; b_1, b_2, \ldots, b_n)$.
- It is easy to check that the new way is as desired. When swapping, the number of blocks in rows and columns does not affect each other.

5. Uniqueness Proof:
- For $n = 1$, the solution is trivial.
- Assume it works for any $i < n$. For the case $n$, without loss of generality, let the first row's number be $n$ and the first column's number be $1$.
- Remove these two lines, and we obtain the uniqueness for $n-1$.
- Each line we remove, which is full of red squares, is also unique for any possible way. This completes the proof.

6. Final Calculation:
- The number of ways to arrange the rows is $n!$ and the number of ways to arrange the columns is also $n!$.
- Therefore, the total number of ways is $2 \times (n!)^2$.

$$2 \times \left| \left\{ (r_1, r_2, \cdots, r_n; c_1, c_2, \cdots, c_n) \mid 1 \leq r_i \neq r_j \leq n, 1 \leq c_i \neq c_j \leq n, 1 \leq i, j \leq n \right\} \right| = 2 \cdot (n!)^2$$

The final answer is $\boxed{2 \cdot (2018!)^2}$