The 71st question:
Answer: Take 1009 squares in the first column and color them red, which can be done in $\mathrm{C}_{2018}^{1009}$ ways. Without loss of generality, assume the first 1009 squares in the first column are red.

In the second column, suppose exactly $\mathrm{k}(0 \leq \mathrm{k} \leq 1009)$ of the first 1009 squares are red, then exactly $1009-\mathrm{k}$ of the last 1009 squares are red. In this case, there are $\mathrm{C}_{1009}^{\mathrm{k}} \cdot \mathrm{C}_{1009}^{1009}-\mathrm{k}=\left(\mathrm{C}_{1009}^{\mathrm{k}}\right)^{2}$ ways to color the second column.
Assume the coloring of the second column is already determined. Next, consider the coloring of the third column.
In a certain row, if the first and second columns are both red, then the third column must be blue (there are $\mathrm{k}$ such rows); if the first and second columns are both blue, then the third column must be red (there are $\mathrm{k}$ such rows); if the first and second columns are one red and one blue, then the third column can be either red or blue (there are $2018-2 \mathrm{k}$ such rows). Since the third column already has $\mathrm{k}$ red squares and $\mathrm{k}$ blue squares, we need to select $1009-\mathrm{k}$ squares from the remaining $2018-2 \mathrm{k}$ squares to color red, and the remaining $1009-\mathrm{k}$ squares to color blue. There are $\mathrm{C}_{2018-2 \mathrm{k}}^{1009}$ ways to do this.
When the coloring of the third column is determined, the coloring of the fourth column is uniquely determined. Therefore,
$$
M=C_{2018}^{1009} \cdot\left[\sum_{k=0}^{1009}\left(C_{1009}^{k}\right)^{2} \cdot C_{2018-2 k}^{1009-k}\right]=C_{2018}^{1009} \cdot\left[\sum_{k=0}^{1009}\left(C_{1009}^{k}\right)^{2} \cdot C_{2 k}^{k}\right]
$$

Notice that $2018=2 \times 1009$, and 1009 is a prime number. Let $\mathrm{p}=1009$, then
$$
M=C_{2 p}^{\mathrm{p}} \cdot\left[\sum_{\mathrm{k}=0}^{\mathrm{p}}\left(\mathrm{C}_{\mathrm{p}}^{\mathrm{k}}\right)^{2} \cdot C_{2 \mathrm{k}}^{\mathrm{k}}\right]
$$

Since $C_{2 p}^{p}=2 \cdot C_{2 p-1}^{p-1} \equiv 0(\bmod 2)$, it follows that $M \equiv 0(\bmod 2)$.
According to Lucas' theorem, $\mathrm{C}_{2 \mathrm{p}}^{\mathrm{p}} \equiv 2(\bmod \mathrm{p})$, and
$$
\sum_{k=0}^{\mathrm{p}}\left(C_{\mathrm{p}}^{\mathrm{k}}\right)^{2} \cdot \mathrm{C}_{2 \mathrm{k}}^{\mathrm{k}}=\sum_{\mathrm{k}=1}^{\mathrm{p}-1}\left(\mathrm{C}_{\mathrm{p}}^{\mathrm{k}}\right)^{2} \cdot \mathrm{C}_{2 \mathrm{k}}^{\mathrm{k}}+1+\mathrm{C}_{2 \mathrm{p}}^{\mathrm{p}} \equiv 0+1+2 \equiv 3(\bmod \mathrm{p})
$$

Thus, $M \equiv 2 \times 3=6(\bmod p)$. Therefore, $M \equiv 6(\bmod 2 p)$.
In conclusion, the remainder when $M$ is divided by 2018 is 6.
The 71st question: In a $2018 \times 4$ grid, each cell is to be colored either red or blue, with each cell being colored exactly one color, such that in each row the number of red and blue cells is the same, and in each column the number of red and blue cells is also the same. Let $\mathrm{M}$ be the number of coloring methods that satisfy the above requirements. Find the remainder when $\mathrm{M}$ is divided by 2018.