Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999

Theorem
Let $n$ be an integer which has at least $3$ digits when expressed in decimal notation.

Let the digits of $n$ be divided into groups of $3$, counting from the right, and those groups added.

Then the result is equal to a multiple of $999$ $n$ is divisible by $999$.

Proof
The mistake is either and conversely or equal to $999$, since $999 \, 999$ is an easy counterexample.

Here we will show that the result is equal to a multiple of $999$ $n$ is divisible by $999$.

Write $n = \displaystyle \sum_{i \mathop = 0}^k a_i 10^{3 i}$, where $0 \le a_i < 1000$.

This divides the digits of $n$ into groups of $3$.

Then the statement is equivalent to:
 * $999 \divides n \iff 999 \divides \displaystyle \sum_{i \mathop = 0}^k a_i$

This statement is true since: