4. Among the 21 numbers $8,9, \cdots, 28$, the probability that the sum of three randomly selected different numbers is exactly divisible by 9 is $\qquad$
4. $\frac{15}{133}$.

Among these 21 numbers, the remainders when divided by 9 are $-1, 0, 1$ each appearing three times, and the remainders $\pm 2, \pm 3, \pm 4$ each appearing twice.

When the selected numbers are grouped by their remainders into an unordered array $(x, y, z)$, they can be categorized into three types for discussion.
(1) None of the selected numbers have a remainder of 0.
Listed as follows: $(-1,-1,2)$ has 6 combinations; $(-1, -2,3)$ has 12 combinations; $(-1,-3,4)$ has 12 combinations; $(-2, -2,4)$ has 2 combinations; $(-2,-3,-4)$ has 8 combinations; $(-1, -4,-4)$ has 3 combinations.

By the symmetry of the signs of the remainders, there are a total of 86 combinations.
(2) The remainders contain only one 0.
There are $3 \times(3 \times 3+2 \times 2 \times 3)=63$ combinations.
(3) All remainders are 0, which is only 1 combination.
Therefore, the number of ways to select numbers that are exactly divisible by 9 is 150.
Thus, the required probability is $\frac{150}{\mathrm{C}_{21}^{3}}=\frac{15}{133}$.