Let n be an odd positive integer.Let M be a set of $n$ positive integers ,which are 2x2 different. A set $T$ $\in$ $M$ is called "good" if the product of its elements is divisible by the sum of the elements in $M$, but is not divisible by the square of the same sum. Given that $M$ is "good",how many "good" subsets of $M$ can there be?
1. **Define the set \( M \) and the conditions for a "good" subset:**
   Let \( M = \{a_1, a_2, \dots, a_n\} \) where \( n \) is an odd positive integer and the elements \( a_i \) are pairwise different positive integers. A subset \( T \subseteq M \) is called "good" if the product of its elements is divisible by the sum of the elements in \( M \), but not divisible by the square of the same sum.

2. **Claim:**
   There are at most \( 2^{n-1} \) good subsets of \( M \).

3. **Proof:**
   Let \( A \) be a good subset of \( M \). If \( M \setminus A \) is also good, then we have:
   \[
   \sum_{a_i \in M} a_i \mid \prod_{a_i \in A} a_i \quad \text{(1)}
   \]
   and
   \[
   \sum_{a_i \in M} a_i \mid \prod_{a_i \in M \setminus A} a_i \quad \text{(2)}
   \]
   Multiplying (1) and (2), we get:
   \[
   \left(\sum_{a_i \in M} a_i\right)^2 \mid \left(\prod_{a_i \in A} a_i \cdot \prod_{a_i \in M \setminus A} a_i\right) = \prod_{a_i \in M} a_i
   \]
   However, this contradicts the fact that \( M \) is good, as \( M \) being good implies that \( \left(\sum_{a_i \in M} a_i\right)^2 \nmid \prod_{a_i \in M} a_i \). Therefore, at least one of \( A \) and \( M \setminus A \) must be good, which proves the claim.

4. **Achievability of \( 2^{n-1} \) good subsets:**
   To show that \( 2^{n-1} \) is achievable, consider \( n = 2k + 1 \) and let \( p \) be a prime such that \( p > \frac{n(n-1)}{2} \). Define the elements of \( M \) as follows:
   \[
   a_i = pi \quad \text{for} \quad 1 \leq i \leq n-1
   \]
   and
   \[
   a_n = p^{k+1} - p \cdot \frac{n(n-1)}{2}
   \]
   Then, the sum of the elements in \( M \) is:
   \[
   a_1 + a_2 + \dots + a_n = p \left(1 + 2 + \dots + (n-1)\right) + a_n = p \cdot \frac{(n-1)n}{2} + a_n = p^{k+1}
   \]
   Therefore, all subsets with more than or exactly \( k+1 \) elements are good, and there are exactly \( 2^{n-1} \) such subsets.

\(\blacksquare\)

The final answer is \( \boxed{ 2^{n-1} } \)