When $n = 2018$, B has a winning strategy; when $n = 2019$, A has a winning strategy.

First, consider an $n \times n$ board where the top and bottom of the board are connected, and the left and right are connected. Since for all $1 \le a, b \le n$, only one piece can be placed among $\{(nx + a, ny + b) \mid x, y \in \mathbb{Z}\}$, we can replace the coordinate plane with such a board, where the cell $(a, b)$ of the board represents the set of points $\{(nx + a, ny + b) \mid x, y \in \mathbb{Z}\}$ in the coordinate plane.

(We adopt the convention that the rows of the board are numbered from $1$ to $n$ from bottom to top, and the columns are numbered from $1$ to $n$ from left to right)

(1) First we prove that when $n$ is even, $B$ has a winning strategy.

Pair up all the cells of the $n \times n$ board: cell $(a, 2t)$ is paired with cell $(a + 2, 2t + 1)$ (here the addition and subtraction of coordinates are all taken $(\bmod\ n)$). If, whenever $A$ places a piece, $B$ places a piece at the cell paired with the cell where that piece was placed, then $A$'s piece each time lies in a new pair, and $B$'s piece completes that pair, so $B$ always has a place to move, and hence $B$ has a winning strategy.

(2) Next we prove that when $n$ is odd, $A$ has a winning strategy. Let $n = 2k + 1$, and define the following sets:

i) When $2 \mid k$, let $S = \{4r + 1, 4r + 2 \mid 0 \le r < \frac{k}{2}\}$, $T = \{4r + 3, 4r + 4 \mid 0 \le r < \frac{k}{2}\}$

ii) When $2$ does not divide $k$, let $S = \{4r + 2, 4r + 3 \mid 0 \le r < \frac{k-1}{2}\} \cup \{2k\}$, $T = \{4r + 4, 4r + 5 \mid 0 \le r < \frac{k-1}{2}\} \cup \{1\}$

It is not hard to see that $S \cup T$ contains exactly all the integers from $1$ to $2k$, and that $T$ is the set formed by adding $2$ to every number in $S$.

Now, pair up the cells of the $n \times n$ board: for $1 \le a \le n$, if $a$ is odd, pair the cell $(a, b): b \in S$ with the cell $(a + 1, b + 2)$; if $a$ is even, pair the cell $(a, b): b \notin T$ with the cell $(a + 1, b - 2)$. From the relationship between $S$ and $T$ described above, we see that all cells except $(1, n)$ are exactly paired up one by one. Therefore $A$ has the following strategy: first place a piece at cell $(1, n)$, and whenever $B$ places a piece, $A$ places a piece at the cell paired with the cell where that piece was placed. It is not hard to see that this is a winning strategy for $A$. This completes the proof.