Find all $f:\mathbb{R}\rightarrow \mathbb{R}$ continuous functions such that $\lim_{x\rightarrow \infty} f(x) =\infty$ and $\forall x,y\in \mathbb{R}, |x-y|>\varphi, \exists n<\varphi^{2023}, n\in \mathbb{N}$ such that 
$$f^n(x)+f^n(y)=x+y$$
1. **Initial Assumptions and Definitions**:
   - We are given a continuous function \( f: \mathbb{R} \rightarrow \mathbb{R} \) such that \(\lim_{x \rightarrow \infty} f(x) = \infty\).
   - For all \( x, y \in \mathbb{R} \) with \( |x - y| > \varphi \), there exists \( n < \varphi^{2023} \) such that \( f^n(x) + f^n(y) = x + y \).

2. **Claim**: The only such continuous function is \( f(x) = x \) for all \( x \in \mathbb{R} \).

3. **Verification**:
   - Clearly, \( f(x) = x \) satisfies the given conditions. We need to show that this is the only function that satisfies these conditions.

4. **Fixed Point Argument**:
   - Suppose \( f \) has no fixed point. Then \( f(x) > x \) for all \( x \in \mathbb{R} \) or \( f(x) < x \) for all \( x \in \mathbb{R} \) due to continuity.
   - Take any \( x, y \) with \( |x - y| > \varphi \). There exists \( n \in \mathbb{N} \) such that \( f^n(x) + f^n(y) = x + y \).
   - This implies \( f^n(x) - x = -(f^n(y) - y) \). By continuity, there exists \( r \in [x, y] \) such that \( f^n(r) = r \).

5. **Contradiction**:
   - If \( f(x) > x \) for all \( x \in \mathbb{R} \) or \( f(x) < x \) for all \( x \in \mathbb{R} \), then \( f^n(r) = r \) is impossible. Hence, \( f \) must have a fixed point, say \( a \).

6. **Interval Analysis**:
   - Consider the interval \( [a - \varphi, a + \varphi] \). Any \( r \) outside this interval satisfies \( |r - a| > \varphi \) and thus has some \( n \) such that \( f^n(r) + f^n(a) = r + a \), implying \( f^n(r) = r \).

7. **Monotonicity and Injectivity**:
   - If \( f(x) = f(y) \) for some \( x, y \) not in the interval \( I = [a - \varphi, a + \varphi] \), then \( x = y \) because \( f^N(x) = x \) and \( f^N(y) = y \).
   - Since \( f \) is injective and continuous over \( (a + \varphi, \infty) \), it follows that \( f \) is monotonous over it. Similarly, \( f \) is monotonous over \( (-\infty, a - \varphi) \).

8. **Behavior at Infinity**:
   - Since \( \lim_{k \rightarrow \infty} f(k) = \infty \), \( f \) is increasing over \( (a + \varphi, \infty) \).
   - If \( f(x) > x \) for some \( x \in (a + \varphi, \infty) \), then the sequence \( \{f^i(x)\}_{i \geq 1} \) is unbounded, contradicting \( f^N(x) = x \). So \( f(x) \leq x \) over this interval.

9. **Upper Bound Argument**:
   - Let \( M \) be an upper bound over all \( f(x) \) with \( x < a + \varphi \). Pick \( x > M \) and consider \( \{f^i(x)\}_{i \geq 1} \). Each term of this sequence is always less than \( x \), unless \( f(x) = x \). Since \( f^N(x) = x \), it must hold that \( f(x) = x \).

10. **Conclusion**:
    - By extending the argument, we see that \( f(x) = x \) for all \( x \in \mathbb{R} \).

\(\blacksquare\)

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