Let $f(n)$ be the number of ways to express positive integer $n$ as a sum of positive odd integers. 
Compute $f(n).$
(If the order of odd numbers are different,  then it is considered as different expression.)
1. **Define the function \( f(n) \):**
   Let \( f(n) \) be the number of ways to express the positive integer \( n \) as a sum of positive odd integers, where the order of summands matters.

2. **Base cases:**
   - For \( n = 1 \), the only way to express 1 as a sum of positive odd integers is \( 1 \) itself. Thus, \( f(1) = 1 \).
   - For \( n = 2 \), the only way to express 2 as a sum of positive odd integers is \( 1 + 1 \). Thus, \( f(2) = 1 \).

3. **Recursive relation:**
   - Consider the first number in the sum. It can be any positive odd integer less than or equal to \( n \). Let the first number be \( k \), where \( k \) is an odd integer.
   - If the first number is \( k \), then the remaining sum is \( n - k \). The number of ways to express \( n - k \) as a sum of positive odd integers is \( f(n - k) \).
   - Therefore, we can write:
     \[
     f(n) = f(n-1) + f(n-3) + f(n-5) + \cdots
     \]
   - Notice that \( f(n-2) = f(n-3) + f(n-5) + f(n-7) + \cdots \). Thus, we can rewrite the above equation as:
     \[
     f(n) = f(n-1) + f(n-2)
     \]
   - This is the Fibonacci recurrence relation.

4. **Conclusion:**
   - Since \( f(n) \) satisfies the Fibonacci recurrence relation and the initial conditions \( f(1) = 1 \) and \( f(2) = 1 \), it follows that \( f(n) \) is the \( n \)-th Fibonacci number.

\(\blacksquare\)

The final answer is \( \boxed{ f(n) } \) is the \( n \)-th Fibonacci number.