 An infinite set $S$ of positive numbers is called thick, if in every interval of the form $\left [1/(n+1),1/n\right]$ (where $n$ is an arbitrary positive integer) there is a number which is the difference of two elements from $S$. Does there exist a thick set such that the sum of its elements is finite?

Proposed by [i]Gábor Szűcs[/i], Szikszó
1. **Define the set \( A_n \):**
   Let 
   \[
   A_n = \left\{\frac{1}{n^2} \right\} \cup \left\{\frac{1}{n^2} - \frac{1}{x} : x \in \mathbb{Z}, n^2 < x \leq (n+1)^2 \right\}.
   \]
   This set includes the element \(\frac{1}{n^2}\) and all elements of the form \(\frac{1}{n^2} - \frac{1}{x}\) where \(x\) is an integer such that \(n^2 < x \leq (n+1)^2\).

2. **Verify the differences:**
   Consider the interval \(\left[\frac{1}{n+1}, \frac{1}{n}\right]\). We need to show that there is a number in this interval which is the difference of two elements from \(A_n\).
   - The elements of \(A_n\) are of the form \(\frac{1}{n^2}\) and \(\frac{1}{n^2} - \frac{1}{x}\).
   - The differences between these elements are:
     \[
     \left(\frac{1}{n^2} - \frac{1}{x}\right) - \frac{1}{n^2} = -\frac{1}{x}.
     \]
   - For \(x\) in the range \(n^2 < x \leq (n+1)^2\), \(\frac{1}{(n+1)^2} \leq \frac{1}{x} < \frac{1}{n^2}\).
   - Therefore, \(\frac{1}{n^2} - \frac{1}{x}\) will cover the interval \(\left[\frac{1}{(n+1)^2}, \frac{1}{n^2}\right]\).

3. **Construct the set \( A \):**
   Define the set \( A \) as:
   \[
   A = \bigcup_{n \in \mathbb{Z}_+} A_n.
   \]
   This set \( A \) is thick because for every interval \(\left[\frac{1}{n+1}, \frac{1}{n}\right]\), there is a number in \(A\) which is the difference of two elements from \(A\).

4. **Sum of elements in \( A_n \):**
   Calculate the sum of elements in \(A_n\):
   \[
   \text{Sum of elements in } A_n < \frac{1}{n^2} + (2n+1) \times \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right).
   \]
   Simplify the expression:
   \[
   \frac{1}{n^2} + (2n+1) \times \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right) = \frac{1}{n^2} + (2n+1) \times \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right).
   \]
   \[
   = \frac{1}{n^2} + (2n+1) \times \left( \frac{2n+1}{n^2(n+1)^2} \right).
   \]
   \[
   = \frac{1}{n^2} + \frac{(2n+1)^2}{n^2(n+1)^2}.
   \]
   \[
   = \frac{1}{n^2} + \frac{4n^2 + 4n + 1}{n^2(n+1)^2}.
   \]
   \[
   < \frac{1}{n^2} + \frac{4n^2 + 4n + 1}{n^4} = \frac{1}{n^2} + \frac{4}{n^2} + \frac{4}{n^3} + \frac{1}{n^4}.
   \]
   \[
   < \frac{5}{n^2}.
   \]

5. **Sum of elements in \( A \):**
   Since \(\sum_{n} \frac{1}{n^2} < \infty\), the sum of elements in \(A\) is finite.

Conclusion:
\[
\boxed{\text{Yes}}
\]