A can balance the crates with the following strategy, with or without B's cooperation.
While the two are alternating turns, A should:
1. If the boxes are balanced, put 1 coin in one box, 2 coin in the other.
2. If The boxes are imbalanced, put 3 coins in the lighter box.
Once B runs out of coins, A can balance the crates with the addition of B's 3 coins:
1. If the crates are balanced, alternate putting 3 coins in each crate.
2. If the crates are imbalanced, A can put B's 3 coins in the lighter crate.
2.1 Then put 2 coins into the lighter box, 1 coin into the heavier box, and use the remaining coins to once again alternate.
Proof:
Claim: When A and B are alternating turns, A can guarantee the difference between the crates to always be 1 after A's move. After B's move, the difference between the crates will not exceed 4.
Base case: starting with 0 coins in each in each case, A will split the coins 2/1 in each crate, which satisfies the claim above. B has four moves after: 3/0, 0/3, 2/1, 1/2. Each then results in: 5/1, 2/4, 4/2, 3/3. In all 4 cases, the difference is at most 4 apart.
Induction:
At the beginning of A's turn, the two crates are at most 4 apart. Because the total number of coins is even, the difference can only be 0, 2, or 4.
Case 0: Same as base case.
Case 2: The crates must have a coin distribution of (3n+2/3n+4) before A's turn because the sum of coins has to be a multiple of 6. A will put 3 in the lighter crate (3n+5/3n+4), which makes the crates only one apart.
Case 4: The crates must have a coin distribution of (3n+1/3n+5) before A's turn because the sum of coins has to be a multiple of 6. A will put 3 in the lighter crate (3n+4/3n+5), which makes the crate only one apart.
At the beginning of B's turn the crates is always 1 apart. After B's turn, the difference in coins between the crates will not exceed 4. Say the crates are at (n/n+1) when B starts:
Case (3/0): B puts 3 coins into the lighter box resulting in (n+3/n+1). The difference is 2, less than 4.
Case (0/3): B puts 3 coins into the heavier box resulting in (n/n+4), the difference is 4.
Case (1/2): B puts 1 into the lighter box, 2 into the heavier one resulting in (n+1/n+3), difference is 2.
Case (2/1) B puts 2 into the lighter box, 1 into the heavier one resulting in (n+2/n+2), which makes the crates balanced.
After B runs out of coins A will have 6m+3 coins, and gain another 3 coins from B. As shown above after B's turn, the two crates are at most 4 apart and the difference can only be even: 0, 2, or 4.
Balanced case: A can balance the crates by alternating putting 3 coins in each crate.
Imbalanced case: A can determine the difference between the two crates by putting the 3 coins from B into the lighter crate. The crates are either in (3n+1/3n+5) or (3n+2/3n+4) and become (3n+4/3n+5) or (3n+5/3n+4). Following this, the crates can be balanced by putting 2 coins into the lighter crate and 1 coin into the heavier crate resulting in (3n+6/3n+6). A now has 6m coins left, which can be distributed evenly by alternating coin deposits in the crates.
Note that m can be 0 for the boxes to be balanced. A can balance the crates by having at least as many coins as B.