What smallest number should be added to 16809 so that the sum is completely divisible by 41?

We need the smallest number to add to 16809 so that the result is divisible by 41.

First find the remainder when 16809 is divided by 41:

$41 \times 410 = 16810$
So, $16809$ is $1$ less than $16810$.

Hence, remainder = $40$ (or equivalently, $-1$ modulo 41).

To make it divisible, we need to add:
$41 - 40 = 1$

Correct Answer: 1