Solution:

$S_n$ is the sum of $g$ over the even numbers $2, 4, \ldots, 2^n.$ Among these $2^{n-1}$ numbers, exactly $2^{n-1-i}$ are divisible by $2^i$ but not $2^{i+1}$ (so have $g = 2^i$) for each $1 \le i \le n - 1,$ and the single number $2^n$ has $g = 2^n.$ Hence $$S_n = \sum_{i=1}^{n-1} 2^i \cdot 2^{n-1-i} + 2^n = (n-1) 2^{n-1} + 2^n = (n+1) 2^{n-1}.$$

If $n$ is even, then $n + 1$ is odd and the exponent $n - 1$ is odd, so $S_n$ has an odd number of factors of $2$ and cannot be a perfect square. If $n$ is odd, then $2^{n-1}$ is already a perfect square, so $S_n$ is a perfect square exactly when $n + 1$ is.

For odd $n,$ the number $n + 1$ is even, and the greatest even perfect square at most $1000$ is $900 = 30^2.$ Thus the greatest valid $n \lt 1000$ is $n = 899.$