Example 2 There are three piles of stones. Each time, A moves one stone from one pile to another, and A can receive a reward from B for each move, which is equal to the difference between the number of stones in the pile to which A moves the stone and the number of stones in the pile from which A moves the stone. If this difference is negative, A should return this amount of reward to B (if unable to pay, A can owe it temporarily). It is known that at a certain moment, all the stones are in their initial piles. Try to find the maximum reward A can earn at this moment.

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【Analysis】Due to the uncertainty of A's operations, it is necessary to start from the whole and establish a "substitute" for the increase or decrease of A's reward each time, which must be simple to calculate.
Solution A's reward is 0.
In fact, the three piles of stones can be imagined as three complete graphs (each stone in the same pile is connected by a line segment, and stones in different piles are not connected).

Each time A operates, a vertex $u$ is moved out from a complete graph $A_{1}$ and into another complete graph $A_{2}$, forming two new complete graphs $A_{1}^{\prime}$ and $A_{2}^{\prime}$. The reward is the difference in the number of vertices between $A_{2}^{\prime}$ and $A_{1}$, which is the difference in the sum of the number of edges between $A_{1}^{\prime}$ and $A_{2}^{\prime}$ and the sum of the number of edges between $A_{1}$ and $A_{2}$.

Thus, A's reward for each operation is the change in the number of edges in the graph.
Since it eventually returns to the initial state, A's reward is 0.
【Comment】By establishing the above substitute, the conclusion becomes clear at a glance.