7. Color each cell of a 2018 by 4 grid either red or blue, such that the number of red cells equals the number of blue cells in each row, and the number of red cells equals the number of blue cells in each column. Let $m$ be the total number of coloring methods that satisfy the above requirements. Find the remainder when $m$ is divided by 2018.
7. A coloring method that satisfies the conditions is called a "pattern", and all patterns form a set $M$.

Use $(i, j)$ to denote the cell in the $i$-th row and $j$-th column, and consider the row number $i$ modulo 2018, i.e., the 2018-th row can be seen as the 0-th row. For a pattern $A$, an operation $T$ can be performed: re-color, changing the color of $(i, j)$ to the color of $(i-1, j)$, i.e.,

after the operation, the 1st row becomes the 2nd row, the 2nd row becomes the 3rd row, ..., and the 2018-th row becomes the 1st row, while the columns remain unchanged. The result of the operation $T$ on pattern $A$, denoted as $T(A)$, also satisfies the problem's requirements and is therefore in $M$.

The operation $T$ is a one-to-one mapping from the set $M$ to itself. Each element $A$ in $M$ returns to itself after 2018 operations $T$, i.e., $T^{(2018)}(A)=A$.

Consider the period of $A$, i.e., the smallest positive integer $d=d(A)$ such that $T^{(d)}(A)=A$, and the $d$-element set
$$
R(A)=\left\{A, T(A), \cdots, T^{(d-1)}(A)\right\}
$$
is called the orbit of $A$.
Clearly, the orbits generated by other elements in $R(A)$ are also $R(A)$.

For two elements $A_{1}$ and $A_{2}$ in $M$, if $R\left(A_{1}\right) \cap R\left(A_{2}\right) \neq \varnothing$, then $R\left(A_{1}\right)=R\left(A_{2}\right)$, i.e., any two orbits are either identical or disjoint, and $M$ can be seen as the disjoint union of several orbits.

The period $d=d(A)$ must be a divisor of 2018, i.e., one of $1, 2, 1009, 2018$. Since each column has exactly 1009 red cells out of 2018 cells, $T^{(1009)}(A) \neq A$ (otherwise, the red cells can be paired, leading to a contradiction in parity), so the period $d(A)$ is either 2 or 2018.

If $d(A)=2$, then all odd-numbered rows of $A$ are colored the same way, and all even-numbered rows are colored the same way but differently from the odd-numbered rows. The first row of $A$ has $\mathrm{C}_{4}^{2}=6$ coloring methods, and the first row determines all other rows. Thus, there are exactly 6 elements in $M$ with a period of 2, and the periods of the other elements are all 2018.

Considering $M$ as the disjoint union of orbits, there are three orbits of length 2 and several orbits of length 2018, so the number of elements $m=|M|$ leaves a remainder of 6 when divided by 2018.