Sum collision from two lists of numbers
You can find a solution with an algorithm that requires about $2^{24}$ operations and $2^{22}$ space, and requires only $N=2^{22}$ entries in each list.  (If you have more than that in each list, ignore the rest and use only the first $2^{22}$ entries.)  Here's how.

Step 1. Build a list $L$ of all of the pairs $(a,b)$ such that $a+b \equiv 0 \bmod 2^{22}$.

By the birthday paradox, we expect there will be about $N^2/2^{22} \approx 2^{22}$ such pairs $(a,b)$, since there are $N^2$ possible pairs and each pair has a $1/2^{22}$ chance of satisfying the condition.  To find them, build a sorted list of the values $a \bmod 2^{22}$, a sorted list of the values $-b \bmod 2^{22}$, and merge the sorted lists.  Or, use a hash table.

Step 2. Look for two pairs $(a_1,b_1)$ and $(a_2,b_2)$ in $L$ that have the same sum modulo $2^{64}$.

By the birthday paradox, we expect such a solution should exist with high probability, since we have about $2^{22}$ entries in $L$ (and thus about $2^{43}$ candidate solutions) and the condition holds with probability $1/2^{42}$.  (Why $1/2^{42}$?  Well, their sum is guaranteed to match in the low 22 bits, by the way we constructed $L$.  So, we just need their high 42 bits to match too.)