Problem 2110

We are tasked with finding the smallest positive number $\lambda$ such that for any 12 points on the plane $P_1, P_2, \ldots, P_{12}$ (which can overlap), if the distance between any two of them does not exceed 1, then $\sum_{1 \le i < j \le 12} |P_iP_j|^2 \le \lambda$.

Let $O$ be an arbitrary point, and let $a_i = \overrightarrow{OP_i}$. We have:
$$\sigma := \sum_{1 \le i < j \le 12} |P_iP_j|^2 = \sum_{1 \le i < j \le 12} (a_i - a_j)^2 = 12 \sum_{i=1}^{12} a_i^2 - \left( \sum_{i=1}^{12} a_i \right)^2 \le 12 \sum_{i=1}^{12} a_i^2.$$

By the Universal Covering Problem, we can cover the set $\{P_1, \ldots, P_{12}\}$ with a circle of radius $\frac{1}{\sqrt{3}}$. Choosing $O$ to be the center of this circle gives $a_i^2 \le \frac{1}{3}$. Therefore,
$$\sigma \le 12 \sum_{i=1}^{12} a_i^2 \le 12 \cdot 12 \cdot \frac{1}{3} = 48.$$

Hence, the smallest positive number $\lambda$ is:
$$\lambda = 48.$$

The answer is: $\boxed{48}.$