Since $f(2) \ge f(1) + f(1) = 2 \cdot f(1)$ and $f(2) = 0$ then $f(1) \le 0$. Therefore, $f(1) = 0$. On the other hand, from $f(3) > 0$ and $f(3) - f(2) - f(1) \in \{0, 1\}$ we have $f(3) = 1$. Putting $m = 1$ in the condition $f(m+n) - f(n) - f(m) \in \{0, 1\}$, we obtain $f(n+1) - f(1) - f(n) = f(n+1) - f(n) \in \{0, 1\}$ from which follows that $f(n+1) \ge f(n)$ for all positive integer $n$. That is, $f$ is increasing.

Now we will prove by induction that $f(3n) \ge n$. The case $n = 1$ trivially holds. Assume that $f(3n) \ge n$ and we have to see that $f(3n + 3) \ge n + 1$. Indeed, $f(3n + 3) - f(3n) - f(3) \ge 0$. So, $f(3n + 3) \ge f(3n) + f(3) \ge n + 1$. On account that $f(6042) = 2014$ then $f(3n) = n$ for $1 \le n \le 2014$. Otherwise, if for some $n \in \{1, 2, 3, \dots, 2014\}$ we have a strict inequality, then it is not possible to obtain $f(6042) = 2014$.

Since $f(2014) = f(3 \cdot 671 + 1)$ if we see that $f(3n + 1) = n$ for all $n \in \{1, 2, 3, \dots, 2014\}$, then $f(2014) = 671$. Finally, we will prove that $f(3n+1) = n$. Indeed, $f(3n + 1) \ge f(3n) + f(1) = f(3n) = n$. On the other hand, we have $3n + 1 = f(9n + 3) \ge f(6n + 2) + f(3n + 1) \ge 3 \cdot f(3n + 1)$ from which follows $f(3n + 1) \le n + \frac{1}{3} < n + 1$. Hence, $f(3n + 1) = n$.