Solution:

The answer is (C). Let us call $a_{1}, a_{2}, \ldots, a_{53}$ the piles ordered in increasing order of height.

We observe that the game ends. On each move the total number of coins decreases by 1, and if there are 51 coins in total then the game has already ended because there are at least two piles with 0 coins.

Certainly, if the game has not already been lost, $a_{i}$ will have fewer coins than $a_{j}$ if $i<j$: if this were not so, since on each move the number of coins decreases by 1, there would be a moment in which the two piles have the same number of coins, but this implies that the game has ended, which is absurd.

Moreover, the only configuration in which every move leads the player to lose (let us call it the decisive configuration) is the one in which the heights of the piles are $0,1,2, \ldots, 52$, in some order. Indeed, if the difference in height between $a_{i+1}$ and $a_{i}$ were at least 2 for some $i$, then we could remove a coin from $a_{i+1}$; if $a_{1}$ had more than 0 coins then we could remove a coin from it.

Therefore the pile $a_{i}$ at the end of the game will have $i-1$ coins for every $i \in\{1,2, \ldots, 53\}$. The number $N$ of coins that will be removed to reach the decisive configuration is equal to the difference between the number of coins present initially and the number of coins in the final configuration, that is
$$N=2020-(0+1+\cdots+52)=2020-\frac{52 \cdot 53}{2}=642$$
Since we can write $N=4 \cdot 160+2$ and since there are 4 players, when the decisive configuration is reached it will be Claudio's turn, so Claudio loses.