Let the sum of the digits of a 1995-digit number divisible by 9 be $a$. We then add the digits of $a$, and let the result be $b$. What can the sum of the digits of $b$ be?
The value of $a$ is the largest when the original number consists of 1995 nines, so $a \leq 1995 \cdot 9 = 17955$. According to this, $a$ is either a four-digit number or a five-digit number starting with 1; therefore, $b \leq 1 + 4 \cdot 9 = 37$. It is clear that the sum of the digits of $b$ is less than $3 + 9 = 12$, yet it is definitely positive, since we started from a non-zero number (if we consider zero as a 1995-digit number, then the sum of the digits of $b$ could also be zero).

Since the original number was divisible by 9, the sum of its digits is also divisible by 9. Thus, $a$ is, $b$ is, and the sum of the digits of $b$ is also divisible by 9. This, together with the previously established limits, means that the sum of the digits of $b$ is 9.

Based on the work of Csaba Megyeri (Nagykanizsa, Batthyány L. Gymnasium, 2nd grade)