Does there exist a positive odd integer $n$ so that there are primes $p_1$, $p_2$ dividing $2^n-1$ with $p_1-p_2=2$?
1. **Identify the problem requirements**: We need to find a positive odd integer \( n \) such that there exist two primes \( p_1 \) and \( p_2 \) dividing \( 2^n - 1 \) with \( p_1 - p_2 = 2 \).

2. **Understand the conditions for \( p_1 \) and \( p_2 \)**: For \( p_1 \) and \( p_2 \) to divide \( 2^n - 1 \), the orders of 2 modulo \( p_1 \) and \( p_2 \) must divide \( n \). Additionally, \( p_1 \) and \( p_2 \) must be twin primes (i.e., \( p_1 - p_2 = 2 \)).

3. **Check quadratic residues**: For \( 2 \) to be a quadratic residue modulo \( p_1 \) and \( p_2 \), we need \( 2^{\frac{p_1-1}{2}} \equiv 1 \pmod{p_1} \) and \( 2^{\frac{p_2-1}{2}} \equiv 1 \pmod{p_2} \). This implies that \( p_1 \) and \( p_2 \) must be congruent to \( 1 \) or \( 7 \) modulo \( 8 \).

4. **Find suitable twin primes**: By checking small twin primes, we find that the smallest pair satisfying \( p_1 \equiv 1 \pmod{8} \) and \( p_2 \equiv 7 \pmod{8} \) is \( 71 \) and \( 73 \).

5. **Verify the orders**: 
   - For \( p_1 = 71 \), we need to check the order of 2 modulo 71. The order must divide \( 70 \). Since \( \nu_2(70) = 1 \), the order of 2 modulo 71 divides \( 35 \).
   - For \( p_2 = 73 \), we need to check the order of 2 modulo 73. Since \( 73 \mid 2^9 - 1 \), the order of 2 modulo 73 is 9, which is odd.

6. **Construct \( n \)**: Since the orders of 2 modulo 71 and 73 are both odd, we can take \( n \) to be the product of these orders. Therefore, \( n \) is an odd integer.

Conclusion:
\[
\boxed{\text{True}}
\]