Let $S_1, S_2, \ldots, S_{100}$ be finite sets of integers such that their intersection is not empty. For every non-empty subset $T$ of $\{S_1, S_2, \ldots, S_{100}\}$, the size of the intersection of the sets in $T$ is a multiple of the number of sets in $T$.

We want to determine the least possible number of elements that are present in at least $50$ of these sets.

### Analysis

Let $n_T = | \bigcap_{S_i \in T} S_i |$, where $T$ is any non-empty subset of the $100$ sets. According to the problem, $n_T$ is a multiple of $|T|$.

To solve this problem, consider:

1. Simplify the Problem: We need to ensure that the intersection of any subset of the provided sets contains an integer and it must also satisfy the condition that $|T|$ divides $n_T$.

2. Constructing an Example:
- Suppose we take an arbitrary integer $c$ that belongs to each $S_i$. This ensures that the intersection of any collection of these sets is not empty, providing the condition that their intersection contains at least one integer.
- Choose $c$ to be part of an arithmetic progression with a common difference that is a multiple of the number of sets involved, ensuring the condition of $|T| \mid n_T$ is satisfied.

3. Estimation:
- Suppose there is an integer $a$ present in exactly $50$ of the sets, i.e., $a$ is included in sets forming a combination of $50$.
- For $\binom{100}{50}$ combinations of choosing 50 sets from 100 sets, if $a$ is the common element, then each such $50$ set combination includes $a$.

4. Count the Minimum Elements:
- Consider each integer to be in exactly 50 sets.
- Thus, for each of the combinations $\binom{100}{50}$, we need an integer present in all 50, giving:
$$50 \times \binom{100}{50}$$
- This product ensures that each combination of $50$ sets from the total $100$ sets has at least 50 in common (and subsequently multiples for larger sets).

Thus, the least possible number of integers that can be in at least 50 sets:
$$\boxed{50 \cdot \binom{100}{50}}$$