Let $ N$ denote the set of natural numbers. Let $ \phi: N\rightarrow N$ be a bijective function and assume that there exists a finite limit
\[ \lim_{n\rightarrow\infty}\frac{\phi(n)}{n}\equal{}L.
\] What are the possible values of $ L$?
1. **Assume there exists \( N \) such that \( \phi(n) > n \) for all \( n \geq N \):**

   - If \( \phi(n) > n \) for all \( n \geq N \), then \( \phi(n) \) takes values greater than \( N \) for all \( n \geq N \).
   - This implies that the values \( \phi(n) \) for \( n \geq N \) are all greater than \( N \), leaving not enough values to fill the set \( \{1, 2, \ldots, N\} \).
   - Therefore, for any \( N \), there must exist some \( n \geq N \) such that \( \phi(n) \leq n \).
   - Hence, \( \liminf_{n \to \infty} \frac{\phi(n)}{n} \leq 1 \).

2. **Assume there exists \( N \) such that \( \phi(n) < n \) for all \( n \geq N \):**

   - Let \( m = \max \{\phi(k) \mid 1 \leq k \leq N\} \).
   - Clearly, \( m \geq N \) because \( \phi \) is a bijection and must cover all natural numbers.
   - However, if \( \phi(n) < n \) for all \( n \geq N \), then \( \max \{\phi(k) \mid 1 \leq k \leq m+1\} = m \), which is a contradiction because \( \phi \) must map to all values up to \( m+1 \).
   - Therefore, for any \( N \), there must exist some \( n \geq N \) such that \( \phi(n) \geq n \).
   - Hence, \( \limsup_{n \to \infty} \frac{\phi(n)}{n} \geq 1 \).

3. **Conclusion:**

   - Since \( \liminf_{n \to \infty} \frac{\phi(n)}{n} \leq 1 \) and \( \limsup_{n \to \infty} \frac{\phi(n)}{n} \geq 1 \), if the limit \( \lim_{n \to \infty} \frac{\phi(n)}{n} \) exists, it must be equal to 1.
   - A trivial example is \( \phi(n) = n \) for \( n \geq 1 \).
   - Another example is \( \phi(2n-1) = 2n \) and \( \phi(2n) = 2n-1 \) for \( n \geq 1 \).

\(\blacksquare\)

The final answer is \( \boxed{ 1 } \).