Let $A$ be a ten-billion digit, positive number divisible by nine. The sum of the digits of $A$ is $B$, the sum of the digits of $B$ is $C$. What is the sum of the digits of $C$?
It is clear that the sum of the digits of a positive number is also positive; it is also well known that if a number is divisible by nine, the sum of its digits is also divisible by nine. Consequently, the number $B$ is divisible by nine, as is $C$, and indeed the sum of the digits of $C$, $D$, is also divisible by nine.

Since $A$ has ten billion digits, the sum of its digits cannot be greater than the sum of ten billion nines: $B \leq 9 \cdot 10^{10}$. In other words, $B$ is a number with at most 11 digits, which cannot consist of 11 nines, as that would be greater than $9 \cdot 10^{10}$. From this, however, we conclude that $C$ is definitely less than $9 \cdot 11$, which is 99. Thus, $D$ is the sum of two digits, at least one of which is less than 9; therefore, $C \leq 9+8=17$. Since the only positive number less than 17 and divisible by 9 is 9, it follows that $D=9$.

Note. The relatively large number of incorrect solutions arose from the confusion between the condition that $A$ has ten billion digits and the condition that $A$ is of the order of ten billion, i.e., 11-digit.