This is an application of the Pigeonhole Principle. The idea is that since there are $n$ possible remainders when an integer is divided by $n$ that at least two of the $n + 1$ integers in the set $\{x_0, x_1, \ldots, x_n\}$ must have the same remainder when divided by $n$. If they have the same remainder when divided by $n$, their difference is divisible by $n$.
The possible remainders when an integer is divided by $n$ are $0, 1, 2, \ldots, n - 1$. If you have $n$ integers, then it is possible for each of them to have a different remainder when divided by $n$. However, if you have $n + 1$ integers, at least two of them must have the same remainder when divided by $n$. Hence, in the set $\{x_0, x_1, \ldots, x_n\}$, there are two numbers $x_i$ and $x_j$, with $i \neq j$, that have the same remainder when they are divided by $n$. Thus, there exist $k, m, r \in \mathbb{N}$, with $0 \leq r \leq n - 1$, such that
\begin{align*}
x_i & = kn + r\\
x_j & = mn + r
\end{align*}
If we take their difference, we obtain
$$x_i - x_j = (kn + r) - (mn + r) = kn - mn = (k - m)n$$
Therefore, $x_i - x_j$ is divisible by $n$.
For four or more digit numbers, we do have a divisibility test by $13$ owing to $1001$ being divisible by $13$ ie, $1000$ leaving remainder $-1$. One checks the alternating sum of blocks of three consecutive digits. Eg, $\overline{abcdefgh}$ is divisible by $13$ if $\overline{ab}-\overline{cde}+\overline{fgh}$ is.
So we check how many $n$ there for which following is multiple of $13$ $$\overline{3}-\overline{0a0}+\overline{b03}=\overline{(b-1)(10-a)6}$$
We have $130+26=156$ one such number $\Rightarrow b=2$, $a=5$.
Others can be found by adding/subtracting $130$.
Best Answer
Since $0+0 < a + b < c + c = 2c$ and $ c | a + b $, hence $ a + b = c$.
Since $ a < b$, hence $ 2a < a+b =c < 2 b $.
Since $0+ b < a + c < b + 2b $, and $b|a+c$ hence $ a+c = 2b$.
Solving the system of equations, we get that $b=2a, c=3a$. It's easy to check that $\{a, 2a, 3a\}$ satisfies the conditions.