Let $$N = \sum^{512}_{i=0}i {512 \choose i}.$$ What is the greatest integer $a$ such that $2^a$ is a divisor of $N$?
1. We start with the given sum:
   \[
   N = \sum_{i=0}^{512} i \binom{512}{i}
   \]

2. We use the known combinatorial identity:
   \[
   \sum_{i=0}^{n} i \binom{n}{i} = n \cdot 2^{n-1}
   \]
   This identity can be derived by considering the number of ways to choose a committee of any nonzero size from \( n \) members and to label one of its members as the "president".

3. Applying this identity to our problem where \( n = 512 \):
   \[
   \sum_{i=0}^{512} i \binom{512}{i} = 512 \cdot 2^{511}
   \]

4. We simplify the expression:
   \[
   512 \cdot 2^{511} = 2^9 \cdot 2^{511} = 2^{520}
   \]

5. Therefore, the greatest integer \( a \) such that \( 2^a \) is a divisor of \( N \) is:
   \[
   a = 520
   \]

The final answer is \( \boxed{520} \)