This answer seems too simple to be correct, so I'm probably misunderstanding some of the rules, but I claim that I can get
>! 11594 gold

as follows:
>! Black Turn 1: Place bishops on d4 and i9, and attack the rooks on e5 and h8  
>! White Turn 1: e5 and h8 are only protected by rooks and queens, so rooks will defend. The bishop on e5 is captured by a rook from either f5 or e6 and the bishop on h8 is captured by a rook from either g8 or h7.  
>! Black Turn 2: Place knights on the two squares that were just vacated by White's rooks. These attack the queens for victory. (f5/e6 threaten g7 and g8/h7 threaten f6)  
>! Costs: Killing both queens nets 12000 gold. Killing them on step two loses me 400 gold, and placing two units on each of the first two turns costs another 6.  

Is this optimal?
>! With this understanding of the rules, yes. Clearly, there is no way to kill either queen on turn 1, and both sacrifices are required in order to get close enough.

A variation, to distinguish from the other answer:
>! Bd4xe5 can be replaced with either Nc8xe7 or Nh3xg5, and likewise Bi9xh8 with either Nj5xh6 or Ne10xf8.

EDIT: Ah! A surprise! How will I get through these pawns?
>! Black 1: Na7xc8 Nl6xj5 (these knights are paralyzed, but they die anyway so I don't care)  
>! White 1: Since I get to resolve the pawn conflicts, I choose d7xc8 i6xj5  
>! Black 2: Nd7xf6 Ni6xg7 for the win, with same rewards as before!  
>! And if knights are forbidden I'll need three turns (11388 gold):  
>! Black 1: Bl3xk4 Ba10xb9  
>! White 1: j5xk4 c8xb9  
>! Black 2: Bj5xi6 Bc8xd7  
>! White 2: Rh6xi6 Re7xd7 (Rooks have priority over pawns)  
>! Black 3: Bh6xg7 Be7xf6