$P(x)$ is a polynomial with real coefficients such that $P(a_1) = 0, P(a_{i+1}) = a_i$ ($i = 1, 2,\ldots$) where $\{a_i\}_{i=1,2,\ldots}$ is an infinite sequence of distinct natural numbers. Determine the possible values of degree of $P(x)$.
1. **Initial Consideration**:
   - We are given that \( P(x) \) is a polynomial with real coefficients such that \( P(a_1) = 0 \) and \( P(a_{i+1}) = a_i \) for \( i = 1, 2, \ldots \), where \( \{a_i\}_{i=1,2,\ldots} \) is an infinite sequence of distinct natural numbers.
   - Clearly, \( P \) cannot be a constant polynomial because if \( P \) were constant, then \( P \) would be identically zero, and the sequence \( (a_i) \) would not consist of distinct numbers (it would be constant and equal to 0). Therefore, \( \deg P \geq 1 \).

2. **Rational Coefficients**:
   - Since \( P \) takes more than its degree natural values at natural arguments, it will have rational coefficients. This follows from the fact that a polynomial that takes integer values at integer points must have rational coefficients (by the Lagrange interpolation polynomial for a set of \( \deg P + 1 \) natural arguments).

3. **Degree Consideration**:
   - We need to consider the implications of the degree of \( P \). Suppose \( \deg P \geq 2 \). By a result from the 2009 Master of Mathematics competition, given a polynomial \( f(x) \) with rational coefficients and degree \( d \geq 2 \), the sequence of sets \( f^0(\mathbb{Q}), f^1(\mathbb{Q}), \ldots \) defined as \( f^0(\mathbb{Q}) = \mathbb{Q} \) and \( f^{n+1}(\mathbb{Q}) = f(f^n(\mathbb{Q})) \) for \( n \geq 0 \), leads to \( f^{\omega}(\mathbb{Q}) = \bigcap_{n=0}^{\infty} f^n(\mathbb{Q}) \) being a finite set.

4. **Contradiction for \( \deg P \geq 2 \)**:
   - If \( \deg P \geq 2 \), then the set \( \{a_i \mid i = 1, 2, \ldots\} \subseteq P^{\omega}(\mathbb{Q}) \) would be finite, which contradicts the given condition that \( \{a_i\} \) is an infinite sequence of distinct natural numbers.

5. **Conclusion**:
   - Therefore, the only possible degree for \( P \) that satisfies the given conditions is \( \deg P = 1 \). Indeed, a polynomial like \( P(x) = x - 1 \) works, as it generates the sequence \( a_i = i \).

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