On the blackboard, we have $15$ integers with residue $0$ modulo $3$ and $30$ integers with residue $1$ modulo $3$. If, in a certain round, Berta and Clara cross out numbers that have the same residue modulo $3$ as the number crossed out by Anna, then they have removed either $0+0+0$ or $1+1+1$ modulo $3$.

In both cases, the sum does not change modulo $3$ and therefore remains divisible by $3$. Since $15$ and $30$ are multiples of $3$, it is possible for Berta and Clara to always choose the same residue as Anna. Therefore, Anna cannot stop Clara from reaching her goal if Clara has Berta's help.

However, if Berta chooses in the first round a residue modulo $3$ that is different from the one chosen by Anna, they have crossed out $0+1$ or $1+0$ modulo $3$. Therefore, Clara's only choices for the sum of the remaining numbers after the first round are $1$ or $2$ modulo $3$. In both cases, the sum is not divisible by $3$. Therefore, Clara has failed in her goal already in the first round if Berta plays uncooperatively.

(Richard Henner) ☐