Chinese Girls' Mathematical Olympiad

Proof Method 1:
(1) For $k = 1, 2, \dots, 2023$, let the coordinates of $P_k$ be $(x_k, y_k)$, and denote $u_k = x_k + y_k$, $v_k = x_k - y_k$. Let $D = \max_{1 \le i < j \le 2023} d(P_i, P_j)$. Then, for any $1 \le i, j \le 2023$,
$$|u_i - u_j| = |(x_i - x_j) + (y_i - y_j)| \le |x_i - x_j| + |y_i - y_j| = d(P_i, P_j) \le D.$$
Thus, the $u_1, u_2, \dots, u_{2023}$ fall within some interval $[a, a+D]$. Similarly for $v_1, v_2, \dots, v_n$. For $k, l = 1, 2, \dots, 44$, consider the region
$$H_{k,l} = \left\{ \left( \frac{u+v}{2}, \frac{u-v}{2} \right) \mid a + \frac{k-1}{44}D \le u \le a + \frac{k}{44}D, \ b + \frac{l-1}{44}D \le v \le b + \frac{l}{44}D \right\}.$$
If $P_i, P_j \in H_{k,l}$, let $U = u_i - u_j, V = v_i - v_j$, then $-\frac{D}{44} \le U, V \le \frac{D}{44}$, we have
$$\begin{aligned} d(P_i, P_j) &= |x_i - x_j| + |y_i - y_j| = \left| \frac{u_i + v_i}{2} - \frac{u_j + v_j}{2} \right| + \left| \frac{u_i - v_i}{2} - \frac{u_j - v_j}{2} \right| \\ &= \left| \frac{U + V}{2} \right| + \left| \frac{U - V}{2} \right| \in \left\{ \pm \frac{U + V}{2} \pm \frac{U - V}{2} \right\} = \{U, -U, V, -V\}. \end{aligned}$$
Hence, $\min_{1 \le i < j \le 2023} d(P_i, P_j) \le d(P_i, P_j) \le \frac{D}{44}$, thus $\lambda \ge 44$.

(2) Construction
Consider the point set
$$M = \{(x, y) \in \mathbb{Z}^2 \mid x, y \text{ have the same parity, } |x + y| \le 44, |x - y| \le 44\}.$$
This set contains $45^2 = 2025$ points. Selecting any 2023 points, the distance $d(P_i, P_j) = |x_i - x_j| + |y_i - y_j|$ is even and greater than 0, i.e., $d(P_i, P_j) \ge 2$. On the other hand,
$$d(P_i, P_j) = |x_i - x_j| + |y_i - y_j| \le 88.$$
Thus,
$$\lambda = \frac{\max_{1 \le i < j \le 2023} d(P_i, P_j)}{\min_{1 \le i < j \le 2023} d(P_i, P_j)} \le 44,$$
and by (1) $\lambda = 44$.

Proof Method Two:
Here is another proof that $\lambda \ge 44$. Assume the coordinates of the 2023 points are $P_k(x_k, y_k)$, $k = 1, 2, \dots, 2023$, and without loss of generality, assume $x_1 \le x_2 \le \dots \le x_{2023}$. By the Erdős-Szekeres theorem, among $y_1, y_2, \dots, y_{2023}$, there exists either an increasing or decreasing subsequence of length $\lceil\sqrt{2023}\rceil = 45$. Assume we have $y_{i_1} \le y_{i_2} \le \dots \le y_{i_{45}}$, with indices $i_1 < i_2 < \dots < i_{45}$. Then,
$$d(P_{i_{45}}, P_{i_1}) = (x_{i_{45}} - x_{i_1}) + (y_{i_{45}} - y_{i_1}) = \sum_{k=1}^{44} (x_{i_{k+1}} - x_{i_k}) + (y_{i_{k+1}} - y_{i_k}) = \sum_{k=1}^{44} d(P_{i_{k+1}}, P_{i_k}).$$
If we have a decreasing subsequence $y_{i_1} \ge y_{i_2} \ge \dots \ge y_{i_{45}}$, then
$$d(P_{i_{45}}, P_{i_1}) = (x_{i_{45}} - x_{i_1}) - (y_{i_{45}} - y_{i_1}) = \sum_{k=1}^{44} (x_{i_{k+1}} - x_{i_k}) - (y_{i_{k+1}} - y_{i_k}) = \sum_{k=1}^{44} d(P_{i_{k+1}}, P_{i_k}).$$
In both cases, we have $d(P_{i_{45}}, P_{i_1}) = \sum_{k=1}^{44} d(P_{i_{k+1}}, P_{i_k})$. Therefore,
$$\lambda = \frac{\max_{1 \le i < j \le 2023} d(P_i, P_j)}{\min_{1 \le i < j \le 2023} d(P_i, P_j)} \ge \frac{d(P_{i_{45}}, P_{i_1})}{\min_{k=1,2,...,44} d(P_{i_{k+1}}, P_{i_k})} \ge 44.$$