11. (20 points) Let the set $A=\left\{z_{1}, z_{2}, \cdots, z_{2024}\right\}$, where $z_{k} \in \mathbf{C}, z_{k} \neq 0 (k=1,2, \cdots, 2024)$. If for any positive integers $1 \leqslant k, m \leqslant 2024$, we have
$$
\begin{aligned}
\frac{z_{m}}{z_{k}} 、 z_{m} z_{k} & \in A, \text { find the set } \\
B= & \left\{\left(z_{x}, z_{y}\right) \mid \operatorname{Re}\left(z_{x}+z_{y}\right)+\operatorname{Im}\left(z_{x}+z_{y}\right)>0, \right. \\
& 1 \leqslant x \leqslant 2024, 1 \leqslant y \leqslant 2024, \\
& x 、 y \in \mathbf{N}\}
\end{aligned}
$$

the number of elements $|B|$ in the set $B$.
11. Let $| z_{t} | = \max \left\{ | z_{1} |, | z_{2} |, \cdots, | z_{2024} | \right\}$. Then
$$
\frac{z_{t}}{z_{t}} = 1 \in A, z_{t}^{2} \in A.
$$

Hence, $| z_{t}^{2} | \leqslant | z_{t} |$. Therefore, $| z_{t} | \leqslant 1$.
Similarly, let
$$
| z_{s} | = \min \left\{ | z_{1} |, | z_{2} |, \cdots, | z_{2024} | \right\},
$$

then $| z_{s} | \geqslant 1$.
Thus, $| z_{x} | = 1 (x = 1, 2, \cdots, 2024)$.
Let $0 = \arg z_{1}0, \right. \\
1 \leqslant x \leqslant 2024, 1 \leqslant y \leqslant 2024, \\
x, y \in \mathbf{N}\}, \\
C = \left\{ (z_{x}, z_{y}) \mid \operatorname{Re}(z_{x} + z_{y}) + \operatorname{Im}(z_{x} + z_{y}) < 0, 1 \right. \\
\leqslant x \leqslant 2024, 1 \leqslant y \leqslant 2024, \\
x, y \in \mathbf{N}\}, \\
D = \left\{ (z_{x}, z_{y}) \mid \operatorname{Re}(z_{x} + z_{y}) + \operatorname{Im}(z_{x} + z_{y}) = 0, \right. \\
1 \leqslant x \leqslant 2024, 1 \leqslant y \leqslant 2024, \\
x, y \in \mathbf{N}\}.
$$

Then $| B \cup C \cup D | = 2024^{2}$, and $B, C, D$ are pairwise disjoint.
Notice that, for any element $(z_{x}, z_{y})$ in $B$,
$$
\begin{array}{l}
(z_{x+1012}, z_{y+1012}), (z_{x+1012}, z_{y-1012}), \\
(z_{x-1012}, z_{y+1012}), (z_{x-1012}, z_{y-1012})
\end{array}
$$

exactly one of them is in $C$, and vice versa.
Thus, $| B | = | C |$.
$$
\begin{array}{l}
\text{Let } \operatorname{Re}(z_{x} + z_{y}) + \operatorname{Im}(z_{x} + z_{y}) = 0, \text{ then} \\
\cos x \theta + \sin x \theta + \cos y \theta + \sin y \theta \\
= 2 \sqrt{2} \sin \left( \frac{x+y}{2} \theta + \frac{\pi}{4} \right) \cdot \cos \frac{x-y}{2} \theta = 0 \\
\Rightarrow \frac{x+y}{2} \theta + \frac{\pi}{4} = k \pi \text{ or } \frac{x-y}{2} \theta = \frac{(2 m + 1) \pi}{2} \\
\Rightarrow x + y = 2024 k - 506 \text{ or } \\
\quad x - y = 1012(2 m + 1),
\end{array}
$$

where $1 \leqslant x, y \leqslant 2024, x, y, k, m \in \mathbf{Z}$.
Thus, $x + y = 1518$ or $x + y = 3542$ or $x - y = \pm 1012$.
Therefore, $| D | = 4046$.
Hence, $| B | = \frac{2024^{2} - 4046}{2}$
$= 2 \times 1012^{2} - 2023$.