Determine the least positive integer $n{}$ for which the following statement is true: the product of any $n{}$ odd consecutive positive integers is divisible by $45$.
To determine the least positive integer \( n \) for which the product of any \( n \) odd consecutive positive integers is divisible by \( 45 \), we need to ensure that the product is divisible by both \( 3^2 = 9 \) and \( 5 \). 

1. **Divisibility by \( 3^2 = 9 \):**
   - For any set of \( n \) consecutive odd integers, we need at least two of them to be divisible by \( 3 \). 
   - Since every third odd number is divisible by \( 3 \), we need \( n \) to be at least \( 6 \) to ensure that there are at least two multiples of \( 3 \) in the set. This is because \( \left\lfloor \frac{6}{3} \right\rfloor = 2 \).

2. **Divisibility by \( 5 \):**
   - For any set of \( n \) consecutive odd integers, we need at least one of them to be divisible by \( 5 \).
   - Since every fifth odd number is divisible by \( 5 \), we need \( n \) to be at least \( 5 \) to ensure that there is at least one multiple of \( 5 \) in the set. This is because \( \left\lfloor \frac{5}{5} \right\rfloor = 1 \).

3. **Combining the conditions:**
   - To satisfy both conditions simultaneously, we need \( n \) to be at least \( 6 \). This ensures that there are at least two multiples of \( 3 \) and at least one multiple of \( 5 \) in any set of \( n \) consecutive odd integers.

4. **Verification for \( n = 5 \):**
   - Consider the set \( \{17, 19, 21, 23, 25\} \). The product is \( 17 \times 19 \times 21 \times 23 \times 25 \).
   - Here, \( 21 \) is divisible by \( 3 \) and \( 25 \) is divisible by \( 5 \), but there is only one multiple of \( 3 \) and one multiple of \( 5 \). Thus, the product is not guaranteed to be divisible by \( 9 \).

5. **Verification for \( n = 6 \):**
   - Consider any set of \( 6 \) consecutive odd integers. For example, \( \{1, 3, 5, 7, 9, 11\} \).
   - In this set, \( 3 \) and \( 9 \) are multiples of \( 3 \), and \( 5 \) is a multiple of \( 5 \). Thus, the product is divisible by \( 9 \times 5 = 45 \).

Therefore, the least positive integer \( n \) for which the product of any \( n \) odd consecutive positive integers is divisible by \( 45 \) is \( 6 \).

The final answer is \( \boxed{6} \).