European Mathematical Cup

To solve this problem, we need to determine if there is a set $S$ consisting of 2012 positive integers such that:

1. For any two distinct elements $a$ and $b$ in $S$, their greatest common divisor $\text{GCD}(a, b) > 1$.
2. For any three distinct elements $a$, $b$, and $c$ in $S$, their greatest common divisor $\text{GCD}(a, b, c) = 1$.

The challenge is to construct such a set $S$ of 2012 elements.

### Construction of the Set $S$

Consider the prime numbers $p_1, p_2, \ldots, p_{2012}$. We can define each element $a_i$ of the set $S$ as the product of two distinct primes from our chosen set of 2012 primes. Specifically, let:

$$a_i = p_i \cdot p_{2012}$$

for each $i = 1, 2, \ldots, 2011$. Let the final element be:

$$a_{2012} = p_1 \cdot p_2$$

### Verification

1. Pairwise GCD Greater than 1:

For any two distinct elements $a_i = p_i \cdot p_{2012}$ and $a_j = p_j \cdot p_{2012}$ where $i, j < 2012$, we have:
$$\text{GCD}(a_i, a_j) = \text{GCD}(p_i \cdot p_{2012}, p_j \cdot p_{2012}) = p_{2012}$$

For any such $a_i$ and $a_{2012} = p_1 \cdot p_2$, we get:
$$\text{GCD}(a_i, a_{2012}) = \text{GCD}(p_i \cdot p_{2012}, p_1 \cdot p_2)$$

As long as neither $p_1$ nor $p_2$ is used among the first 2011 elements (which it isn't because $i \neq 1, 2$ for $i < 2012$), the GCD will include at least one factor from the product, hence:
$$\text{GCD}(a_i, a_{2012}) = 1$$

2. Triple GCD Equal to 1:

Considering any three distinct elements $a_i$, $a_j$, and $a_k$, where two of them are from the first 2011 elements and one is $a_{2012}$:
$$\text{GCD}(a_i, a_j, a_{2012}) = \text{GCD}(p_i \cdot p_{2012}, p_j \cdot p_{2012}, p_1 \cdot p_2) = 1$$

Since the $a_i$ are pairwise coprime with the additional element $a_{2012}$, the GCD condition with three distinct elements holds, as their triples always share only one prime, thus:
$$\text{GCD}(a_i, a_j, a_k) = 1$$

### Conclusion

By this choice and verification, it demonstrates that such a set $S$ of 2012 elements satisfying all required conditions can be constructed. Therefore, it is indeed possible to have a set $S$ with 2012 elements following these conditions. Hence, the answer is:
$$\boxed{\text{Yes}}$$