From (6,21) the robot can either go to (27,21) or (6,27). Another turn after that, the robot will be at either (48,21), (27, 48), (33, 27) or (6, 33). Because these numbers have no way of ever decreasing, the robot can go to any point (6a+21b,6c+21d) where a, b, c and d are all positive integers, b and c being larger than 0. Hope this answers your question!
(After I saw Rand's solution, it seemed to me that there ought to be a way of streamlining mine a bit. So I've done that. The original version of mine is preserved below in case anyone feels that the improved one is "polluted" by my having seen OP's own answer.)
The smallest possible $a+b$ is
First of all, notice that
There is one ...
Find the prime factorization of the modulus: 3^1*5^1
For each prime number in the factorization, take the prime number reduced by one power times one less than the prime number, and then multiply all of those numbers together. In this case, both prime numbers are raised to the first power, so reducing the power leaves just p^0, or 1. So we have (3-1)(5-1) = ...
This is a special case of a more general problem:
How to find the $N$th positive integer coprime to $k$?
The important thing to know is that
Therefore, to find the $N$th positive integer coprime to $k$, we first
So basically, to solve this general problem you only need to know
and then the rest is just basic arithmetic.