The answer is $2(2018!)^2$. We consider the problem for an $n \times n$ board. Let $r_i$ be the number of red squares in the $i$-th row for $i = 1, \dots, n$ and $c_j$ be the number of red squares in the $j$-th column for $j = 1, \dots, n$. Since $r_i \neq r_j$ for every $i \neq j$, we see that $\{r_1, \dots, r_n\} = \{0, 1, \dots, n\} \setminus \{a\}$ for some number $a$. Similarly, we have $\{c_1, \dots, c_n\} = \{0, 1, \dots, n\} \setminus \{b\}$ for some number $b$. On the other hand, the number of red squares on the board is both $\sum_{i=1}^n r_i$ and $\sum_{j=1}^n c_j$. Thus, we get $a=b$.

Note that if $r_i = 0$ for some $i$, then $c_j \le n-1$ for every $j$ and hence, $a = n$. Moreover, if $r_i = n$ for some $i$, then $c_j \ge 1$ for every $j$ and hence, $a = 0$. Therefore, the missing number is either $0$ or $n$.
If $a = 0$, then $r_i = c_j = n$ for some $i$ and $j$ and removing the $i$-th row and $j$-th column gives us the problem for $(n-1) \times (n-1)$ board with the missing number $n-1$. Similarly, If $a = n$, then $r_i = c_j = 0$ for some $i$ and $j$ and removing the $i$-th row and $j$-th column gives us the problem for $(n-1) \times (n-1)$ board with the missing number $0$. Therefore, at the beginning we can decide on the missing number in $2$ ways but later on it is uniquely determined. We can choose a row and a column to remove in $n \cdot n$ ways. Finally, the answer is
$$2 \cdot n \cdot n \cdot (n-1) \cdot (n-1) \cdots 2 \cdot 2 \cdot 1 \cdot 1 = 2(n!)^2.$$