Determine the smallest positive integer $m$ such that $529^n+m\cdot 132^n$ is divisible by $262417$ for all odd positive integers $n$.
To determine the smallest positive integer \( m \) such that \( 529^n + m \cdot 132^n \) is divisible by \( 262417 \) for all odd positive integers \( n \), we start by factoring \( 262417 \).

1. **Factorize \( 262417 \):**
   \[
   262417 = 397 \times 661
   \]
   Both \( 397 \) and \( 661 \) are prime numbers.

2. **Consider the divisibility conditions:**
   For \( 529^n + m \cdot 132^n \) to be divisible by \( 262417 \), it must be divisible by both \( 397 \) and \( 661 \).

3. **Divisibility by \( 397 \):**
   \[
   529 \equiv 1 \pmod{397}
   \]
   This is because \( 529 = 23^2 \) and \( 23^2 \equiv 1 \pmod{397} \) (since \( 23 \) is a primitive root modulo \( 397 \)).
   Therefore,
   \[
   529^n \equiv 1^n \equiv 1 \pmod{397}
   \]
   Hence,
   \[
   529^n + m \cdot 132^n \equiv 1 + m \cdot 132^n \equiv 0 \pmod{397}
   \]
   This implies:
   \[
   1 + m \cdot 132^n \equiv 0 \pmod{397}
   \]
   Since \( 132 \equiv -1 \pmod{397} \) (as \( 132 = 397 - 265 \)), we have:
   \[
   1 + m \cdot (-1)^n \equiv 0 \pmod{397}
   \]
   For odd \( n \), \( (-1)^n = -1 \), so:
   \[
   1 - m \equiv 0 \pmod{397} \implies m \equiv 1 \pmod{397}
   \]

4. **Divisibility by \( 661 \):**
   \[
   529 \equiv -132 \pmod{661}
   \]
   This is because \( 529 = 23^2 \) and \( 132 = 661 - 529 \).
   Therefore,
   \[
   529^n \equiv (-132)^n \pmod{661}
   \]
   Hence,
   \[
   529^n + m \cdot 132^n \equiv (-132)^n + m \cdot 132^n \equiv 0 \pmod{661}
   \]
   For odd \( n \), \( (-132)^n = -132^n \), so:
   \[
   -132^n + m \cdot 132^n \equiv 0 \pmod{661}
   \]
   This implies:
   \[
   m \cdot 132^n \equiv 132^n \pmod{661} \implies m \equiv 1 \pmod{661}
   \]

5. **Combine the congruences:**
   We have:
   \[
   m \equiv 1 \pmod{397}
   \]
   and
   \[
   m \equiv 1 \pmod{661}
   \]
   By the Chinese Remainder Theorem, since \( 397 \) and \( 661 \) are coprime, there is a unique solution modulo \( 262417 \):
   \[
   m \equiv 1 \pmod{262417}
   \]

Therefore, the smallest positive integer \( m \) that satisfies the conditions is \( m = 1 \).

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