$2 \cdot 46$ Find the smallest positive integer $n$, such that when $n$ different numbers $a_{1}, a_{2}, \cdots, a_{n}$ are chosen from the interval $(1,1000)$, there always exist two numbers $a_{i}, a_{j}, 1 \leqslant i, j \leqslant n$, satisfying the inequality
$$
0<a_{i}-a_{j}<1+3 \sqrt[3]{a_{i} a_{j}} .
$$
［Solution］Since
$$
\begin{aligned}
a_{i}-a_{j} & =\left(\sqrt[3]{a_{i}}-\sqrt[3]{a_{j}}\right)\left(\sqrt[3]{a_{i}^{2}}+\sqrt[3]{a_{i} a_{j}}+\sqrt[3]{a_{j}^{2}}\right) \\
& =\left(\sqrt[3]{a_{i}}-\sqrt[3]{a_{j}}\right)\left[\left(\sqrt[3]{a_{i}}-\sqrt[3]{a_{j}}\right)^{2}+3 \sqrt[3]{a_{i} a_{j}}\right],
\end{aligned}
$$

it follows that the necessary and sufficient condition for (1) to hold is
$$
0<\sqrt[3]{a_{i}}-\sqrt[3]{a_{j}}<1 .
$$

When $a \in[1,1000]$, $\sqrt[3]{a} \in[1,10]$. By the Pigeonhole Principle, when $n \geqslant 11$, there must exist $a_{i}$ and $a_{j}$ such that (2) holds, and thus (1) also holds.

When $n=10$, take $\sqrt[3]{a_{i}}=i, i=1,2, \cdots, 10$, then no two numbers among them satisfy (2) (in this case, (2) becomes an equality), which does not meet the requirements of the problem. Therefore, the smallest positive integer $n=11$.