Will Sawin's idea seems good. Here is a slightly simpler way to get a similar pairwise independent distribution where the maximum load with under a binary choice is still $\Theta(\sqrt{n}).$
Let $\lbrace x_i \rbrace$ be a random sequence of bins that is symmetric under permuting bins and indices so that a random set of about $\sqrt{n}$ balls go into one bin (in fact, at least $\sqrt{n}-1$ go into one bin), and the other balls go into distinct bins. This can be done so that the chance that two balls go into the same bin is $1/n$ so they are pairwise independent.
Choose a random permutation $\pi$ so that for any $i$, $\pi(i)$ is uniform. For example, we can choose $\pi$ to be uniform on $S_n$, or we can choose a random translation mod $n$.
Consider the sequence of pairs $\lbrace (x_i, \pi(x_i)) \rbrace$.
The first coordinates are pairwise independent by the construction of $\lbrace x_i \rbrace$. The second coordinates are pairwise independent by the symmetry of permuting bins and the pairwise independence of $\lbrace x_i \rbrace$. The probability that $x_i = \pi(x_j)$ is $1/n$ by the construction of $\pi$. So, these are pairwise independent.
Since at least $\sqrt{n}-1$ of the pairs are the same $(x,\pi(x))$, at least $(\sqrt{n}-1)/2$ balls have to go into either bin $x$ or bin $\pi(x)$, so the maximum load is $\Theta(\sqrt{n})$.