MAXOR

Let $k$ be the integer satisfying $2^k\leq N<2^{k+1}$.

The numbers $1,2,4,\ldots,2^k$ are all available. Each of them contains exactly one distinct set bit, so we can choose them to set all of the lowest $k+1$ bits. This obtains

$$2^{k+1}-1.$$

Every integer from $1$ to $N$ is smaller than $2^{k+1}$. Therefore, XORing any selection cannot set a higher bit, and no result can exceed $2^{k+1}-1$.

Thus, the answer is one less than the smallest power of two greater than $N$. The time complexity is $O(\log N)$ per test case, and the additional space complexity is $O(1)$.

Solution written by GPT6