Given $1962$ -digit number. It is divisible by $9$. Let $x$ be the sum of its digits. Let the sum of the digits of $x$ be $y$. Let the sum of the digits of $y$ be $z$. Find $z$.
1. Given a $1962$-digit number that is divisible by $9$, we need to find the value of $z$, where $z$ is the sum of the digits of the sum of the digits of the sum of the digits of the original number.
2. Let $x$ be the sum of the digits of the original number. Since the number is divisible by $9$, $x$ must also be divisible by $9$.
3. The maximum possible value of $x$ can be calculated as follows:
   \[
   x \leq 9 \cdot 1962 = 17658
   \]
4. Next, we need to find the sum of the digits of $x$, denoted as $y$. The maximum value of $y$ occurs when $x$ is at its maximum possible value. However, since $x$ is a multiple of $9$, we need to consider the largest multiple of $9$ less than or equal to $17658$.
5. The sum of the digits of $17658$ is:
   \[
   1 + 7 + 6 + 5 + 8 = 27
   \]
   Since $27$ is a multiple of $9$, $y$ must also be a multiple of $9$.
6. The maximum value of $y$ is $27$ (since $27$ is the sum of the digits of $17658$ and is a multiple of $9$).
7. Now, we need to find the sum of the digits of $y$, denoted as $z$. Since $y = 27$, the sum of the digits of $27$ is:
   \[
   2 + 7 = 9
   \]
8. Therefore, $z = 9$.

The final answer is $\boxed{9}$.