Solution:

Answer: $4022$

If $n$ is odd and greater than $1$, then Goliath makes the last move. No matter what two numbers are on the board, Goliath can combine them to make an even number. Hence Goliath has a winning strategy in this case.

Now suppose $n$ is even. We can replace all numbers on the board by their residues modulo $2$. Initially the board reads $1, 0, 1, 0, \ldots, 1, 0$. David combines the rightmost $1$ and $0$ by addition to make $1$, so now the board reads $1, 0, 1, 0, \ldots, 0, 1$. We call a board of this form a "good" board. When it is Goliath's turn, and there is a good board, no matter where he moves, David can make a move to restore a good board. Indeed, Goliath must combine a neighboring $0$ and $1$; David can then combine that number with a neighboring $1$ to make $1$ and create a good board with two fewer numbers.

David can ensure a good board after his last turn. But a good board with one number is simply $1$, so David wins. So David has a winning strategy if $n$ is even. Therefore, the 2011th smallest positive integer greater than $1$ for which David can guarantee victory is the 2011th even positive integer, which is $4022$.