Put $P = m^{2003}n^{2017} - m^{2017}n^{2003}$ , where $m, n \in N$.
a) Is $P$ divisible by $24$?
b) Do there exist $m, n \in N$ such that $P$ is not divisible by $7$?

Let's address each part of the problem step-by-step.

### Part (a): Is \( P \) divisible by \( 24 \)?

1. **Expression for \( P \)**:
   \[
   P = m^{2003}n^{2017} - m^{2017}n^{2003}
   \]

2. **Check divisibility by \( 8 \)**:
   - Consider \( m \) and \( n \) modulo \( 8 \). We need to check if \( P \) is divisible by \( 8 \) for all \( m, n \in \mathbb{N} \).
   - Note that \( m^{2003} \equiv m \pmod{8} \) and \( n^{2017} \equiv n \pmod{8} \) because \( m \) and \( n \) are natural numbers and the exponents are large enough to reduce to the base values modulo \( 8 \).
   - Therefore, \( P \equiv m \cdot n - m \cdot n \equiv 0 \pmod{8} \).

3. **Check divisibility by \( 3 \)**:
   - Similarly, consider \( m \) and \( n \) modulo \( 3 \). We need to check if \( P \) is divisible by \( 3 \) for all \( m, n \in \mathbb{N} \).
   - Note that \( m^{2003} \equiv m \pmod{3} \) and \( n^{2017} \equiv n \pmod{3} \) because \( m \) and \( n \) are natural numbers and the exponents are large enough to reduce to the base values modulo \( 3 \).
   - Therefore, \( P \equiv m \cdot n - m \cdot n \equiv 0 \pmod{3} \).

4. **Combine results using the Chinese Remainder Theorem**:
   - Since \( P \equiv 0 \pmod{8} \) and \( P \equiv 0 \pmod{3} \), by the Chinese Remainder Theorem, \( P \equiv 0 \pmod{24} \).

Thus, \( P \) is divisible by \( 24 \).

### Part (b): Do there exist \( m, n \in \mathbb{N} \) such that \( P \) is not divisible by \( 7 \)?

1. **Expression for \( P \)**:
   \[
   P = m^{2003}n^{2017} - m^{2017}n^{2003}
   \]

2. **Check divisibility by \( 7 \)**:
   - Consider \( m \) and \( n \) modulo \( 7 \). We need to find if there exist \( m, n \in \mathbb{N} \) such that \( P \) is not divisible by \( 7 \).
   - Note that \( m^{2003} \equiv m \pmod{7} \) and \( n^{2017} \equiv n \pmod{7} \) because \( m \) and \( n \) are natural numbers and the exponents are large enough to reduce to the base values modulo \( 7 \).
   - Therefore, \( P \equiv m \cdot n - m \cdot n \equiv 0 \pmod{7} \) if \( m \equiv n \pmod{7} \).

3. **Find counterexample**:
   - Choose \( m \equiv 1 \pmod{7} \) and \( n \equiv 2 \pmod{7} \).
   - Then, \( m^{2003} \equiv 1 \pmod{7} \) and \( n^{2017} \equiv 2 \pmod{7} \).
   - Therefore, \( P \equiv 1 \cdot 2 - 1 \cdot 2 \equiv 2 - 1 \equiv 1 \pmod{7} \).

Thus, there exist \( m, n \in \mathbb{N} \) such that \( P \) is not divisible by \( 7 \).

The final answer is \( \boxed{ P } \) is divisible by \( 24 \) and there exist \( m, n \in \mathbb{N} \) such that \( P \) is not divisible by \( 7 \).