Half the time, the first collector gets less than 5 paintings, so 0, 2 or 4 paintings. We can choose those paintings in 10C0 + 10C2 + 10C4 ways, or
1 + (10)(9)/2! + (10)(9)(8)(7)/4! = 1 + 45 + 210 = 256
ways. The other half of the time, or another 256 times, the first collector gets more than 5 paintings, so doubling the result above, the answer is (2)(256) = 512.
There's a faster way, but it relies on a combinatorial identity you would never need on the GMAT (and this question is not the kind of question you see on the GMAT). The total number of even-sized subsets you can pick from a group is equal to the number of odd-sized subsets you can pick, so we can also count all of the ways to distribute the paintings, and divide by 2. Since we have 2 choices for each of the ten paintings, there are 2^10 ways to distribute them with no restrictions, and 2^10/2 = 2^9 = 512 ways if each collector must get an even number of paintings.
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