For any positive integer n, evaluate $$\sum_{i=0}^{\lfloor \frac{n+1}{2} \rfloor}
 {n-i+1 \choose i}$$
, where  $\lfloor n \rfloor$  is the greatest integer less than or equal to $n$ .

To evaluate the sum 
\[ \sum_{i=0}^{\lfloor \frac{n+1}{2} \rfloor} {n-i+1 \choose i}, \]
we will use a combinatorial argument and relate it to the Fibonacci sequence.

1. **Combinatorial Interpretation:**
   Consider a row of \( n+1 \) balls. We want to count the number of ways to choose \( i \) pairs of consecutive balls, where \( 0 \leq i \leq \left\lfloor \frac{n+1}{2} \right\rfloor \).

2. **Pairs and Unpaired Balls:**
   Each pair of consecutive balls can be treated as a single entity. If we choose \( i \) pairs, we have \( n+1 - 2i \) balls left. These remaining balls can be either paired or unpaired.

3. **Counting Arrangements:**
   The number of ways to arrange \( i \) pairs among \( n-i+1 \) positions (since each pair is treated as a single entity) is given by \( \binom{n-i+1}{i} \).

4. **Summing Over All Possible \( i \):**
   We need to sum this over all possible values of \( i \) from \( 0 \) to \( \left\lfloor \frac{n+1}{2} \right\rfloor \):
   \[ \sum_{i=0}^{\lfloor \frac{n+1}{2} \rfloor} {n-i+1 \choose i}. \]

5. **Recurrence Relation:**
   Consider the total number of ways to arrange \( n+1 \) balls such that each arrangement ends either in an unpaired ball or a pair of balls. This gives the recurrence relation:
   \[ r_{n+1} = r_n + r_{n-1}, \]
   where \( r_k \) denotes the total number of arrangements with \( k \) balls.

6. **Initial Conditions:**
   By brute force computation, we find:
   \[ r_2 = 2 \quad \text{and} \quad r_3 = 3. \]

7. **Fibonacci Sequence:**
   The recurrence relation \( r_{n+1} = r_n + r_{n-1} \) with initial conditions \( r_2 = 2 \) and \( r_3 = 3 \) matches the Fibonacci sequence shifted by one index. Therefore, \( r_{n+1} = F_{n+1} \), where \( F_{n+1} \) is the \((n+1)\)-th Fibonacci number.

8. **Conclusion:**
   Therefore, the sum can be expressed as:
   \[ \sum_{i=0}^{\lfloor \frac{n+1}{2} \rfloor} \binom{n-i+1}{i} = F_{n+1}. \]

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