Find all functions $f : N\rightarrow N - \{1\}$ satisfying $f (n)+ f (n+1)= f (n+2) +f (n+3) -168$ for all $n \in N$
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1. **Define the function \( g(n) \):**
   Let \( g(n) = f(n) + f(n+1) \). The given functional equation is:
   \[
   f(n) + f(n+1) = f(n+2) + f(n+3) - 168
   \]
   This can be rewritten using \( g(n) \) as:
   \[
   g(n) = g(n+2) - 168
   \]

2. **Find the general form of \( g(n) \):**
   From the equation \( g(n+2) = g(n) + 168 \), we can see that \( g(n) \) is an arithmetic sequence with a common difference of 168. Therefore, we can write:
   \[
   g(2n) = 168n + a \quad \text{and} \quad g(2n+1) = 168n + b
   \]
   for some constants \( a \) and \( b \).

3. **Express \( f(n) \) in terms of \( g(n) \):**
   Using the definitions of \( g(n) \), we have:
   \[
   f(2n) + f(2n+1) = 168n + a
   \]
   \[
   f(2n+1) + f(2n+2) = 168n + b
   \]

4. **Subtract the two equations to find a recurrence relation:**
   Subtracting the first equation from the second, we get:
   \[
   f(2n+2) - f(2n) = b - a
   \]
   This implies:
   \[
   f(2n+2) = f(2n) + (b - a)
   \]

5. **Solve the recurrence relation:**
   Let \( d = b - a \). Then:
   \[
   f(2n) = f(0) + nd
   \]
   Let \( f(0) = c \). Therefore:
   \[
   f(2n) = c + nd
   \]

6. **Find \( f(2n+1) \):**
   Using \( f(2n) + f(2n+1) = 168n + a \), we get:
   \[
   f(2n+1) = 168n + a - f(2n)
   \]
   Substituting \( f(2n) = c + nd \):
   \[
   f(2n+1) = 168n + a - (c + nd) = (168 - d)n + a - c
   \]

7. **Check the constraints:**
   Since \( f(n) \) maps to \( \mathbb{N} \setminus \{1\} \), we need:
   \[
   c + nd > 1 \quad \text{and} \quad (168 - d)n + a - c > 1
   \]
   for all \( n \in \mathbb{N} \).

8. **Conclusion:**
   The general solution is:
   \[
   f(2n) = c + nd \quad \text{and} \quad f(2n+1) = (168 - d)n + a - c
   \]
   where \( c \) and \( a \) are constants such that \( c > 1 \) and \( a - c > 1 \).

The final answer is \( \boxed{ f(2n) = c + nd } \) and \( f(2n+1) = (168 - d)n + a - c \).