a)
Let us firstly choose numbers in such a way: $2, 3, 5, \ldots, p_{2016}$, where $p_i$ is a prime number in increasing order. Then we will write down all the possible combinations of subsets $B$ and $C$. Its amount is finite. Let there are $N$ such pairs. Successively we will change the elements in a particular way. And we will build ordered set $A$, at every step we will change some of its elements comparing with the previous step. We will identify the elements as $a_1, a_2, \ldots, a_{2016}$ (at the $0$ step (at the beginning) $a_i = p_i$, $i=1, \ldots, 2016$). Elements of the sets $B$ and $C$ will be determined by their indices. For instance, $B = \{1, 4, 2000\}$ means that the set $B$ includes $a_1, a_4$ and $a_{2000}$ elements of $A$, after their change at the previous step.

Let us look at the sets $A, B, C$ in $k$-th row. Let $p = p_{2016+k}$. Now we multiply all the elements of the set $C$ by $p$, and from the elements of the set $B$ - the first element (with the lowest index $a_j$) we do not multiply by $p$, and the rest of the elements (if the set $B$ contains only one element, we do not have the rest) multiply by $p$. Thus the sum of the elements of the set $B$ is not divisible by $p$, and the sum of the elements of the set $C$ is divisible by $p$, hence the sum of the elements of $B$ is not divisible by the sum of the elements of $C$.

In such a way we will make $N$ steps and show that the obtained set satisfies the conditions.

Let us choose sets $B$ and $C$ arbitrarily (the sets are determined by the sets of indices), let it correspond to $k$-th row of the table. Then from the elements of set $B$ all but one are divisible by $p_{2016+k}$, and among elements of set $C$ all are divisible by it, as is the sum.

It is clear that similarly we can build a corresponding set, that contains arbitrarily finite amount of pairwise distinct elements.

b)
Let us show how such a set can be built for arbitrarily finite amount of elements $n$. Let $N = \left(\frac{n(n+1)}{2}\right)!$. Let us build the corresponding set: $a_i = i \cdot N$, $i=1, \ldots, n$. Let us show that it satisfies the conditions. Consider any subset $C$, sum of its elements $L$ is not greater than $L \le M = (1+2+\ldots+n) \cdot N = \frac{n(n+1)}{2} N$, thus it can be written as $K \cdot N$, where $K \le \frac{n(n+1)}{2}$. The product of any two numbers of the set $A$ is divisible by $N^2$, thus it is divisible by $L$, since $N$ is divisible by any number less than $\frac{n(n+1)}{2}$, thus it is divisible by $K$.