Find all function $f:\mathbb{N} \longrightarrow \mathbb{N}$ such that

forall positive integers $x$ and $y$, $\frac{f(x+y)-f(x)}{f(y)}$  is again a positive integer not exceeding $2022^{2022}$.
1. **Define the problem and initial assumptions:**
   We need to find all functions \( f: \mathbb{N} \to \mathbb{N} \) such that for all positive integers \( x \) and \( y \), the expression 
   \[
   \frac{f(x+y) - f(x)}{f(y)}
   \]
   is a positive integer not exceeding \( 2022^{2022} \). Let \( C = 2022^{2022} \).

2. **Assume \( f \) is a solution and consider scaling:**
   If \( f \) is a solution, then \( Af \) for any constant \( A \) is also a solution. We can assume without loss of generality that \( \gcd(f(1), f(2), \dots) = 1 \). This implies that \( f(1) \) divides all \( f(x) \).

3. **Determine \( f(1) \):**
   Since \( f(1) \) divides all \( f(x) \), we can assume \( f(1) = 1 \). This simplifies our problem because now \( f(x) \) must be such that \( \frac{f(x+y) - f(x)}{f(y)} \leq C \).

4. **Express \( f(x+1) \) in terms of \( f(x) \):**
   Let \( f(x+1) = f(x) + g(x) \), where \( g(x) \) is a bounded function because \( \frac{g(x)}{f(1)} \leq C \). This implies \( g(x) \leq C \).

5. **Analyze the divisibility condition:**
   Swapping \( x \) and \( y \) in the original condition, we get:
   \[
   \frac{f(x+y) - f(x)}{f(y)} \quad \text{and} \quad \frac{f(x+y) - f(y)}{f(x)}
   \]
   Since \( f(x) \mid f(x+1) - f(1) = f(x+1) - 1 \), it follows that \( f(x) \mid g(x) - 1 \).

6. **Boundedness and strict monotonicity:**
   Since \( f \) is strictly increasing and \( g(x) \) is bounded, \( g(x) \) must eventually be 1. This means that for sufficiently large \( n \), \( f(n+1) = f(n) + 1 \).

7. **Inductive step:**
   Fix some \( t \). For large \( n \), we have \( f(n+t) = f(n) + t \). This implies \( t/f(t) \in \mathbb{Z} \). Given \( f(1) = 1 \) and \( f \) is increasing, by induction, \( f(n) \geq n \) for all \( n \).

8. **Conclusion:**
   Since \( f(n) \geq n \) and \( f(n) \) is strictly increasing, the only function that satisfies all conditions is \( f(n) = n \).

\(\blacksquare\)

The final answer is \( \boxed{ f(n) = n } \).