Problem 1911

The largest possible $n$ is $3$.
Note that
$$\begin{aligned} & (3x_i - x_j)(x_i - 3x_j) \ge (1 - x_ix_j)^2 \\ \Leftrightarrow & 3x_i^2 - 10x_ix_j + 3x_j^2 \ge 1 - 2x_ix_j + x_i^2x_j^2 \\ \Leftrightarrow & 3x_i^2 - 6x_ix_j + 3x_j^2 \ge 1 + 2x_ix_j + x_i^2x_j^2 \\ \Leftrightarrow & 3(x_i - x_j)^2 \ge (1 + x_ix_j)^2 \\ \Leftrightarrow & \left| \frac{x_i - x_j}{1 + x_ix_j} \right| \ge \frac{1}{\sqrt{3}}. \end{aligned} \quad (1)$$
WLOG, we may assume $x_1 < x_2 < \dots < x_n$. For $1 \le j \le n$, we can find $\theta_j \in \mathbb{R}$ such that $0 < \theta_j < \frac{\pi}{2}$ and $x_j = \tan \theta_j$. Now for $1 \le i < j \le n$, the inequality (1) is equivalent to
$$\tan(\theta_j - \theta_i) \ge \frac{1}{\sqrt{3}},$$
which means $\theta_j - \theta_i \ge \frac{\pi}{6}$.
If $n \ge 4$, by the pigeonhole principle, there exists $j$ with $1 \le j \le 3$ such that $\theta_{j+1} - \theta_j < \frac{\pi}{6}$. This contradicts (1). Therefore, we must have $n \le 3$.

It is possible to have $n = 3$. For example, we can take $x_j = \tan \frac{(2j-1)\pi}{12}$ for $j = 1, 2, 3$. Following the above arguments, we see that all inequalities are satisfied.