Borrow Low, Lend High

Key idea. Borrow from the account with the low rate and deposit in the account with the high rate. This costs nothing today and earns a sure profit at the end of the year.

Let $x$ be the integer amount deposited in $A$ and $y$ the integer amount deposited in $B$, where a negative amount means borrowing. You start with $0. If $x+y=-c$ with $c>0$, you hold $c$ in idle cash, which earns nothing. Every dollar of idle cash could instead have reduced the $B$-loan, which saves $e^{0.02}>1$ per dollar. So the best strategies are self-financing: $y=-x$.

After one year you close both accounts and receive

This amount is known today, so it is riskless. It is positive exactly when $x>0$, that is, when you borrow from $B$ and deposit in $A$.

The basic arbitrage is one unit: borrow $1 from $B$ and deposit it in $A$. At the end of the year $A$ returns $e^{0.04}\approx 1.04081$. Repaying $B$ costs $e^{0.02}\approx 1.02020$. The locked-in profit is

Remark on scale. The printed statement asks for the largest guaranteed profit, but it sets no limit on position size. Taken literally, that question has no finite answer: using $n$ units multiplies the profit by $n$, so it is unbounded. The meaningful quantity is the profit per unit of the arbitrage. This is also the smallest positive amount you can guarantee under the integer constraint. It is the answer below.