Always good to be concrete, so without loss of generality, let’s say J=99 i.e. 198 players.
Q1: Given 2J players, we need to allocate exactly half of them to NY and the rest to San Francisco, to form two teams for a inter-city contest. We receive the NY fare and SF fare for each player, as 2J (i.e. 198) pairs of integers. How can we minimize the total fares?
Q2(bonus): what if 2J+1 players, and you can decide which city gets an extra player?
— Analysis: how many possible allocations? 2J-choose-J. Very large number of allocations. Something like O(J!) so brute force is impractical.
If for any player the two fares both go up by $9800, it doesn’t change our algorithm at all. Therefore, we only care about the fare difference (N-S) for each player.
— solution: I will assume most players live near SF so SF fares are lower. I tabulate and sort the 198 fare differences “NY fare – SF fare” and suppose indeed at least 99 (half) are positive. Therefore, my “base” is SF i.e. my base allocation is everyone-allocated-to-SF. Next I must pick 99 (half) of them and allocate to NY.
I will avoid the higher values in the table.
I simply pick the lower 99 (half) in the table, because these are the 99 “extra” costs I will incur. I want to minimize the total of these 99 values, whether they are mostly positive or mostly negative.
- Note the highest number in the table indicates the most expensive player for NY relative to SF. If this number is negative, then SF is more expensive than NY for All players so follow the rule in  but notice he is the least expensive players for SF relative to NY. Regardless of positive/negative sign, we want to keep this person in SF .
- Note the lowest number in the table indicates the least expensive player for NY relative to SF. If this number is negative, then SF is more expensive than NY for this player — normal. Regardless, we want to allocate her to NY.
 if proven otherwise (by the data), then we could easily treat NY as base to keep the solution intuitive. Actually even if we don’t change the base, the algorithm still works, albeit unintuitively.
A2: Say J=99 so total 199 players We already picked 99 for NY. Do we pick one more for NY or keep remaining 100 in SF?
Just look at the next lowest value in the table. If positive, then NY is more expensive for him, so keep him in SF.