jmc

Given $g:x\mapsto {\max }_{j:2^j|x}{2}^j$, consider $S_n=g(2)+\cdots +g(2^n)$. Define $S=\{2,4,\dots ,2^n\}$. There are $2^0$ elements of $S$ that are divisible by $2^n$, $2^1-2^0=2^0$ elements of $S$ that are divisible by $2^{n-1}$ but not by $2^n,\dots ,$ and $2^{n-1}-2^{n-2}=2^{n-2}$ elements of $S$ that are divisible by $2^1$ but not by $2^2$. Thus$$\begin{array}{rl}{S}_n& =2^0\cdot 2^n+2^0\cdot 2^{n-1}+2^1\cdot 2^{n-2}+\cdots +2^{n-2}\cdot 2^1\\ & =2^n+(n-1)2^{n-1}\\ & =2^{n-1}(n+1).\end{array}$$Let $2^k$ be the highest power of $2$ that divides $n+1$. Thus by the above formula, the highest power of $2$ that divides $S_n$ is $2^{k+n-1}$. For $S_n$ to be a perfect square, $k+n-1$ must be even. If $k$ is odd, then $n+1$ is even, hence $k+n-1$ is odd, and $S_n$ cannot be a perfect square. Hence $k$ must be even. In particular, as $n<1000$, we have five choices for $k$, namely $k=0,2,4,6,8$. If $k=0$, then $n+1$ is odd, so $k+n-1$ is odd, hence the largest power of $2$ dividing $S_n$ has an odd exponent, so $S_n$ is not a perfect square. In the other cases, note that $k+n-1$ is even, so the highest power of $2$ dividing $S_n$ will be a perfect square. In particular, $S_n$ will be a perfect square if and only if $(n+1)/2^k$ is an odd perfect square. If $k=2$, then $n<1000$ implies that $\frac{n+1}{4}\le 250$, so we have $n+1=4,4\cdot 3^2,\dots ,4\cdot 13^2,4\cdot 3^2\cdot 5^2$. If $k=4$, then $n<1000$ implies that $\frac{n+1}{16}\le 62$, so $n+1=16,16\cdot 3^2,16\cdot 5^2,16\cdot 7^2$. If $k=6$, then $n<1000$ implies that $\frac{n+1}{64}\le 15$, so $n+1=64,64\cdot 3^2$. If $k=8$, then $n<1000$ implies that $\frac{n+1}{256}\le 3$, so $n+1=256$. Comparing the largest term in each case, we find that the maximum possible $n$ such that $S_n$ is a perfect square is $4\cdot 3^2\cdot 5^2-1=\boxed{899}$.