The smallest value $B$ can attain is 2018.

Without loss of generality we can assume that the weight of the lightest ball is equal to 1. If this is not the case, we can divide all the weights by the weight of the lightest ball.

Let us assume that there are at most 2017 balls of each weight appearing in all the bowls. Let $n$ be a non-negative integer such that $a^n$ is the weight of the heaviest ball.

Since there are at most 2017 balls of the weight $a^n$, there exists at least one bowl which does not contain any balls of weight $a^n$, and moreover, all of the balls in that bowl are lighter than $a^n$. Thus, the total weight of balls in that bowl is at most
$$2017 \cdot \sum_{k=0}^{n-1} a^k = 2017 \cdot \frac{a^n - 1}{a - 1} \le 2017 \cdot \frac{a^n - 1}{2017} = a^n - 1 < a^n,$$
which leads to contradiction.

So, there exists at least one weight which appears at least 2018 times.

On the other hand, if we put a ball of weight 1 in each bowl, then all bowls contain the same weight, while the weight 1 appears exactly 2018 times.