1. Define the problem and the given condition:
We are given a collection of finite non-empty sets $S_1, S_2, \ldots, S_n$ such that:
$$\sum_{1 \leq i < j \leq n} \frac{|S_i \cap S_j|}{|S_i||S_j|} < 1.$$
We need to prove that there exist pairwise distinct elements $x_1, x_2, \ldots, x_n$ such that $x_i \in S_i$ for each $i$.

2. Random selection and probability calculation:
Choose a random element $x_i$ from $S_i$ for each $i = 1, 2, \ldots, n$. The probability that the chosen elements $x_i$ and $x_j$ are the same for $i \neq j$ is given by:
$$\frac{|S_i \cap S_j|}{|S_i||S_j|}.$$

3. Summing the probabilities:
The probability that at least one pair $(x_i, x_j)$ are the same (i.e., there exists some $i \neq j$ such that $x_i = x_j$) is at most:
$$\sum_{1 \leq i < j \leq n} \frac{|S_i \cap S_j|}{|S_i||S_j|}.$$
By the given condition, this sum is less than 1:
$$\sum_{1 \leq i < j \leq n} \frac{|S_i \cap S_j|}{|S_i||S_j|} < 1.$$

4. Conclusion using the probabilistic method:
Since the probability that some two $x_i$ and $x_j$ are the same is less than 1, there must exist a choice of $x_1, x_2, \ldots, x_n$ such that all $x_i$ are distinct. This is because if the probability of an event is less than 1, then there exists a non-zero probability that the event does not occur. Hence, there exists a selection where all $x_i$ are distinct.

$\blacksquare$