XI OBM

Note that $B$ will lose if he does not space his tokens widely enough. Call the rows $R$ and $S$, so that $S^n$ denotes the number $n$ in row $S$. Suppose $A$ plays $1$ on $R5$, then $2$ on $R6$. If $B$ plays $1$ on $S5$ and $2$ on $S7$, then he loses, because $A$ swaps rows and plays $3$ on $S6$. However, $B$ does not need to make this mistake, so $A$ can only force a win by always playing in the longer row, until $B$ runs out of space to match him in the shorter row. So $A$ places each token as centrally as possible in one the shortest gap. $B$ must match him. The table

after	1989	1492
1	994	745
2	496	372
3	247	185
4	123	92
5	61	45
6	30	22
7	14	10
8	6	4
9	2	1

Evidently $B$ can always play the first $10$ tokens. But $A$ can fit both tokens $10$ and $11$ into his gap of $2$, whereas $B$ can only fit token $10$. So $B$ loses for more than $10$ tokens, and wins for at most $10$ tokens.

The only way $B$ can lose is if he runs out of space to put his tokens. Clearly he cannot run out of space with the rationals, because there is always a rational between any two given rationals. Similarly, he should not run out of space with all the integers. He just copies $A$ in the other row, so that the tokens numbered $k$ are always placed on the same number in each row.