The function $f: N \to N_0$ is  such that $f (2) = 0, f (3)> 0, f (6042) = 2014$ and  $f (m + n)- f (m) - f (n) \in\{0,1\}$ for all $m,n \in N$. Determine $f (2014)$.  $N_0=\{0,1,2,...\}$
1. Given the function \( f: \mathbb{N} \to \mathbb{N}_0 \) with the properties:
   - \( f(2) = 0 \)
   - \( f(3) > 0 \)
   - \( f(6042) = 2014 \)
   - \( f(m + n) - f(m) - f(n) \in \{0, 1\} \) for all \( m, n \in \mathbb{N} \)

2. We need to determine \( f(2014) \).

3. Consider the equation \( f(6042) - f(2014) - f(4028) \in \{0, 1\} \). Let this difference be \( y \), so:
   \[
   f(6042) - f(2014) - f(4028) = y \quad \text{where} \quad y \in \{0, 1\}
   \]
   Given \( f(6042) = 2014 \), we have:
   \[
   2014 - f(2014) - f(4028) = y
   \]
   Rearranging, we get:
   \[
   f(4028) + f(2014) = 2014 - y
   \]

4. Next, consider the equation \( f(4028) - 2f(2014) \in \{0, 1\} \). Let this difference be \( x \), so:
   \[
   f(4028) - 2f(2014) = x \quad \text{where} \quad x \in \{0, 1\}
   \]
   Rearranging, we get:
   \[
   f(4028) = 2f(2014) + x
   \]

5. Substitute \( f(4028) = 2f(2014) + x \) into the equation \( f(4028) + f(2014) = 2014 - y \):
   \[
   2f(2014) + x + f(2014) = 2014 - y
   \]
   Simplifying, we get:
   \[
   3f(2014) + x = 2014 - y
   \]

6. Rearrange to solve for \( f(2014) \):
   \[
   3f(2014) = 2014 - x - y
   \]

7. Since \( x, y \in \{0, 1\} \), the possible values for \( x + y \) are \( 0, 1, 2 \). For \( 3f(2014) \) to be an integer, \( 2014 - (x + y) \) must be divisible by 3. 

8. Check the divisibility:
   - If \( x + y = 0 \), then \( 2014 \mod 3 = 1 \), not divisible by 3.
   - If \( x + y = 1 \), then \( 2013 \mod 3 = 0 \), divisible by 3.
   - If \( x + y = 2 \), then \( 2012 \mod 3 = 2 \), not divisible by 3.

   Therefore, \( x + y = 1 \).

9. Substitute \( x + y = 1 \) into the equation:
   \[
   3f(2014) = 2014 - 1
   \]
   \[
   3f(2014) = 2013
   \]
   \[
   f(2014) = \frac{2013}{3} = 671
   \]

The final answer is \( \boxed{671} \).