9. In $1,2, \cdots, 2006$, the probability of randomly selecting three numbers that can form an increasing arithmetic sequence is $\qquad$ .
9. $\frac{3}{4010}$.

Three numbers form an increasing arithmetic sequence, denoted as $a, a+d, a+2d$. According to the problem, it must satisfy $a+2d \leqslant 2006, d \leqslant 1002$. For a given $d$, $a$ can take values $1,2, \cdots, 2006-2d$.
Therefore, the number of increasing arithmetic sequences of three numbers is
$$
\sum_{d=1}^{1002}(2006-2d)=1003 \times 1002 \text{. }
$$

The probability that three numbers form an increasing arithmetic sequence is
$$
\frac{1003 \times 1002}{C_{2006}^{3}}=\frac{3}{4010} .
$$