To solve this problem, we need to determine for which natural numbers $n$ there exists a set $S$ of special triples, with $|S| = n$, such that any special triple is bettered by at least one element of $S$.

### Understanding the Definitions

A special triple $(a_1, a_2, a_3)$ is defined as a triple of nonnegative real numbers satisfying $a_1 + a_2 + a_3 = 1$.

A triple $(a_1, a_2, a_3)$ is better than a triple $(b_1, b_2, b_3)$ when exactly two of the following inequalities hold:
$$a_1 > b_1, \quad a_2 > b_2, \quad a_3 > b_3.$$

### Problem Analysis

To construct a set $S$ such that any special triple is bettered by at least one element of $S$, consider the properties of comparisons:

1. Trivial Cases:
- For $n = 1$, if $S$ contains only one special triple, say $(c_1, c_2, c_3)$, there exist other triples such as the same $(c_1, c_2, c_3)$ or others not bettered due to symmetry and equality conditions.
- For $n = 2$ or $n = 3$, it is not possible to construct a set $S$ where every other special triple is worse than at least one in $S$ due to the one-dimensional constraint and the rule of exactly two strict inequalities.

2. **General Case for $n \geq 4$**:
Consider the corners of an equilateral triangle within the space defined by the sum of coordinates being 1. The ideas of domination are achievable by considering triples closer to pure corners, such as:
- $(1, 0, 0)$
- $(0, 1, 0)$
- $(0, 0, 1)$
- Additional points like $(0.5, 0.5, 0)$

These configurations cover enough of the special triples such that for any arbitrary special triple $(a_1, a_2, a_3)$ there is at least one element in $S$ that will better it.

### Conclusion

Based on this construction, it's concluded that for $n \geq 4$, a collection $S$ can be created such that any special triple is bettered by at least one element in $S$.

Thus, the set of natural numbers $n$ for which such a collection $S$ can exist is:
$$\boxed{n \geq 4}$$