(Solution by I. Voronovich.) Rewrite the given equation as
$$f(f(m) + n) = f(n) - f(m) + f(3m) + c, \quad (1)$$
where $c = 2014$; all we need in this solution is that $c$ is even and coprime with 3. Let $f(0) = a$.

1) Setting $m = 0$ we get $f(n+a) = f(n)+c$, then by standard induction to both sides $f(n+ka) = f(n) + kc$, $k \in \mathbb{Z}$; in particular, $f(ka) = a + kc$ and $f(a) = a + c$.

2) Set $m = n$ in (1), then
$$f(n) + n = f(3n) + c = [\text{in view of 1}] = f(3n + a).$$

3) Setting $m = a$ in (1) we get
$$f(n + f(a)) = f(n) - f(a) + f(3a) + c,$$
or, in view of 1) $f(n+c) + c = f(n+c+a) = f(n) - c - a + a + 3c + c$, or $f(n+c) = f(n) + 2c$. Again, by easy induction $f(n+kc) = f(n) + 2kc$.

4) Searching on injectivity: suppose that $f(s) = f(t)$. Then setting $m = s$ and $m = t$ in (1) we obtain $f(3s) = f(3t)$, and thus $f(3^l s) = f(3^l t)$ for all $l \in \mathbb{N}$. Since $(3;c)=1$, there exists an $l$ such that $3^l - 1 = nc$ for some $n \in \mathbb{N}$. Then in view of 3) we have
$$f(3^l s) = f(3^l t) \iff f(s + (3^l - 1)s) = f(t + (3^l - 1)t),$$
or $f(s + nsc) = f(t + ntc)$, or $f(s) + 2nsc = f(t) + 2ntc$, $2nsc = 2ntc$, so $s = t$. Thus $f$ is injective. Since in view of 2) $f(f(n) + n) = f(3n + a)$, we obtain $f(n) + n = 3n + a$, $f(n) = 2n + a$. Easy verification shows that the function is a solution of (1) if and only if $a = c/2$.