When two voters meet, they switch allegiance; might they all ally with the same candidate? 
Let's assume that there are three candidates running in an election.
  Right before the elections (when there is no more propaganda), it is
  forbidden to gather in groups of more than two people to discuss the
  candidates regarding the election.
There are basically three groups of people: group a which would vote
  for candidate A, group b which would vote for candidate B, and group c
  which would vote for candidate C, in such a way that the following is
  true: 0 < |a| < |b| < |c| (If we assume that the groups are sets).
  Each time two people from different groups meet, they change their
  opinions and vote for the third candidate. (So if a person which would
  vote for A meets a person which would vote for B, then after their
  conversation, they both would vote for C).
The question is: can all of the voters (the union of groups a, b,
  c) be persuaded to vote for the same candidate?


Let $s_0 = \newcommand{\tuple}[1]{\left\langle #1 \right\rangle}\tuple{a_0,b_0,c_0}$ be the initial state and $(s_k)_k$ be any sequence of following states. Set $$f(\tuple{a,b,c}) = \tuple{a-b, b-c, c-a} \bmod 3,$$ then $f(s_k) = f(s_0)$ for any $k$.
Proof.
Without loss of generality we can assume that
$$s_k = \tuple{a, b, c} \leadsto \tuple{a-1, b-1, c+2} = s_{k+1}.$$
Then:
\begin{align}
(a-1)-(b-1) &= a-b &&= a-b \pmod 3, \\
(b-1)-(c+2) &= b-c-3 &&= b-c \pmod 3, \\
(c+2)-(a-1) &= c-a+3 &&= c-a \pmod 3,
\end{align}
hence, $f(s_k) = f(s_{k+1})$.
Concluding, it is possible to get $a = b = 0$ only if $a_0-b_0 = 0 \pmod 3$. Moreover, there is a winning strategy if there are at least two different voters: WLOG we have $a \leq b$ (otherwise switch $a$ and $b$). First we obtain $c \geq b$ (trivial). Next we generate:
$\tuple{0, b_0-a_0, c_0+2a_0}$, then $\tuple{2\frac{b_0-a_0}{3},2\frac{b_0-a_0}{3}, c_0+2a_0-1\frac{b_0-a_0}{3}}$, and finally $\tuple{0, 0, a_0+b_0+c_0}$.
I hope this helps ;-)
