How many queens can be placed in a standard $8*8$ chessboard to threat each of them exactly once by other queens?
I found a 10 queen answer. I realized that d'alar'cop's reasoning was slightly off, because threatening is part of the puzzle, so 8 wasn't necessarily a limit. With the 8 inner queens in this position, you can put the final 2 queens in any 2 corners.
Update: I have a proof for a maximum of 10 queens.
Given that 2 queens in a row or a column use 1x2 and 2 queens diagonally use 2x2, it is advantageous to avoid the diagonals.
Put 2 queens in a row. You have now used up 1 row and 2 columns.
Do this again. 2 rows, 4 columns.
Again. 3 rows, 6 columns.
If you do it a 4th time, you will have used all 8 columns, so don't.
Put 2 queens in a column. 5 rows, 7 columns.
Do this again. 7 rows, 8 columns.
You can no longer add any queens, as your columns are used up.
Repeat swapping columns and rows and you end up with 8 rows and 7 columns, so you still cannot add any queens.
Any attempts at using diagonals simply make the problem end faster (exception: a single diagonal, like my displayed answer, gets you to 8 rows, 8 columns).
Therefore, 10 queens is the maximum.
Now as for allowing other pieces, I noticed that there might be extra space from d'alar'cop's 24 queens answer, and I was able to get 2 more queens in.
A queen cannot have a paired queen on the same column if it also has one on the same row. So, queens will pair up every other row and every other column. This will yield 8 queens. There are many configurations (100s probably). Here is one example:
The question didn't strictly state if this matters, but I believe this is a maximum.
If other pieces are allowed, then other pawns can be used to block queens from each other. 24 queens can be placed on the board in this case.