4. A circle is filled with 12 positive integers, each taken from $\{1,2, \cdots, 9\}$ (each number can appear multiple times on the circle). Let $S$ represent the sum of all 12 numbers on the circle. If the sum of any three consecutive numbers on the circle is a multiple of 7, then the number of possible values for $S$ is $\qquad$ kinds.
4.9.

For any three consecutive numbers $a_{k}, a_{k+1}, a_{k+2}$ on a circle, $a_{k}+a_{k+1}+a_{k+2}$ can be 7, 14, or 21.

For any four consecutive numbers on the circle, if they are $a_{k}, a_{k+1}, a_{k+2}, a_{k+3}$, since
$$
a_{k}+a_{k+1}+a_{k+2} \text { and } a_{k+1}+a_{k+2}+a_{k+3}
$$

are both multiples of 7, it must be that
$7 \mid\left(a_{k+3}-a_{k}\right)$.
Thus, $a_{k}$ and $a_{k+3}$ are either equal or differ by 7.
Now, divide the circle into four segments, each with a sum of three consecutive numbers that can be 7, 14, or 21, so the total sum of the four segments can be any of the following:
$$
\{28,35,42,49,56,63,70,77,84\}
$$

There are a total of 9 possible cases.
One way to fill the circle is: first, sequentially fill 12 numbers on the circle:
$$
1,2,4,1,2,4,1,2,4,1,2,4,
$$

with a sum of 28; then, each time change one 1 to 8, or one 2 to 9, each operation increases the total sum by 7. Such operations can be performed 8 times.