No, you can have any number of tiles, so brute force isn't enough. I've implemented a brute force solution to experiment with it, but I need something faster.

You can always score zero with the strategy I describe; you can always score 1 with the strategy you describe. Actually I think you can always score at least ceil(n/2): choose the letter which appears most often above, and lay that down in every place below. If the tiles are alternating then that's the best you can do.

There is a connection with adding, but it's pretty longwinded.