Find all injective function $f: N \to N$ satisfying that for all positive integers $m,n$, we have: $f(n(f(m)) \le nm$
1. Let \( P(m, n) \) be the assertion \( f(nf(m)) \le nm \).

2. Define \( c = f(1) \).

3. Consider \( P(1, n) \):
   \[
   f(nf(1)) \le n \implies f(nc) \le n
   \]

4. Since \( f \) is injective, we can use induction to show that \( f(f(x)) = x \). 

   - Base case: For \( x = 1 \), we have \( f(f(1)) = 1 \).
   - Inductive step: Assume \( f(f(k)) = k \) for some \( k \). Then for \( k+1 \), we have:
     \[
     f(f(k+1)) = k+1
     \]
     This follows from the injectivity of \( f \) and the fact that \( f(f(x)) \le x \).

5. Now consider \( P(x, c) \):
   \[
   f(cf(x)) \le cx
   \]
   Since \( f(f(x)) = x \), we have:
   \[
   f(cf(x)) = f(f(f(x))) = f(x)
   \]
   Thus:
   \[
   f(x) \le cx
   \]

6. Next, consider the inequality \( \frac{f(x)}{x} \le c \):
   \[
   \frac{f(x)}{x} \le c
   \]

7. Using the fact that \( f(f(x)) = x \), we get:
   \[
   \frac{x}{f(x)} = \frac{f(f(x))}{f(x)} \le c \implies \frac{f(x)}{x} \ge \frac{1}{c}
   \]

8. Combining the two inequalities, we have:
   \[
   \frac{1}{c} \le \frac{f(x)}{x} \le c
   \]

9. Consider \( P(x, 1) \):
   \[
   f(f(x)) \le x
   \]
   Since \( f(f(x)) = x \), we have:
   \[
   f(f(x)) = x
   \]

10. Now, consider \( P(x, c) \) again:
    \[
    f(cx) \le cx
    \]
    Since \( f \) is injective and \( f(f(x)) = x \), we have:
    \[
    f(cx) = x
    \]

11. From the above, we conclude that \( f(x) = cx \). Plugging this into the original inequality, we get:
    \[
    f(nf(m)) = f(ncm) = cnm \le nm
    \]
    This implies \( c = 1 \).

12. Therefore, \( f(x) = x \) for all \( x \in \mathbb{N} \).

The final answer is \( \boxed{ f(x) = x } \) for all \( x \in \mathbb{N} \).