Compute the union of two sets between two endpoints minimizing communication complexity
You can do this with $O(K \log N)$ bits of communication on average, where $K$ is the size of $|(A\setminus B) \cup (B\setminus A)|$, assuming you are willing to use a non-interactive protocol and are willing to accept a randomized protocol.

Let $H(\cdot)$ denote a randomized hash function.  Do binary search on $i$ to find the smallest $i$ such that $H(a_1,\dots,a_i) \ne H(b_1,\dots,b_i)$, i.e., (ignoring hash collisions) such that $a_i \ne b_i$.  Then you know that $a_1=b_1$, ..., $a_{i-1}=b_{i-1}$ and $a_i \ne b_i$.  Thus $a$ can send $a_i$ to $B$.  You can use a $O(\log N)$-bit hash function and the probability of a hash collision can be made exponentially small.  This takes $O(\log N)$ iterations of binary search, and in each iteration of binary search you send a $O(\log N)$-bit hash, so in all you send $O((\log N)^2)$ bits of information.  At the end we have sent one of the $K$ elements from $A$ to $B$.  Now you can repeat on the remaining elements (after discarding $a_1,\dots,a_i$ from both $a$ and $b$'s sets).  You repeat at most $K$ times, so this gives a protocol that requires sending $O(K (\log N)^2)$ bits.

You can reduce this to $O(K \log N)$ bits by using a shorter hash.  First do binary search using a 1-bit hash, to find the smallest $i$ such that $H(a_1,\dots,a_i)_1  \ne H(b_1,\dots,b_i)_1$, where $H(\cdot)_b$ denotes the first $b$ bits of the hash $H(\cdot)$.  This  takes $O(\log N)$ iterations.  Then, by using backtracking, you can find the smallest $i$ such that $H(a_1,\dots,a_i)_2  \ne H(b_1,\dots,b_i)_2$ , using on average $O(1)$ iterations.  Then find the smallest $i$ such that $H(a_1,\dots,a_i)_4  \ne H(b_1,\dots,b_i)_4$, and so on, in each stage doubling the length of the hash.  Each stage uses on average $O(1)$ iterations.  If you sum the series of the number of bits sent, it sums  to $O(\log N)$.

Thus, we get an improved protocol that requires sending $O(K \log N)$ bits on average.  It does require interaction between $a$ and $b$ to enable the binary search, and it is randomized.

I don't know whether there is a non-interactive protocol with similar complexity, or whether the protocol can be derandomized to achieve $O(K \log N)$ bits in the worst case rather than in the average case (or rather than an exponentially small probability of sending more than $O(K \log N)$ bits).