The graph $G$ with 2014 vertices doesn’t contain any 3-cliques. If the set of the degrees of the vertices of $G$ is $\{1,2,...,k\}$, find the greatest possible value of $k$.
1. **Restate the problem and notation:**
   Let \( G \) be a graph with \( n = 2014 \) vertices that does not contain any 3-cliques (triangles). The set of degrees of the vertices of \( G \) is \( \{1, 2, \ldots, k\} \). We need to find the greatest possible value of \( k \).

2. **Initial claim and reasoning:**
   We claim that the maximum value of \( k \) is \( \left\lfloor \frac{2 \cdot 2014}{3} \right\rfloor = 1342 \). To justify this, consider a vertex with degree \( k \). All its neighbors must have degrees less than or equal to \( n - k \) because no two of them are connected (as \( G \) has no 3-cliques).

3. **Degree constraints:**
   The neighbors of a vertex with degree \( k \) can have at most \( n - k \) different degrees. The remaining \( n - 1 - k \) vertices can have at most \( n - k - 1 \) different degrees. Therefore, the total number of different degrees must satisfy:
   \[
   1 + (n - k) + (n - k - 1) \geq k
   \]
   Simplifying this inequality:
   \[
   1 + n - k + n - k - 1 \geq k
   \]
   \[
   2n - 2k \geq k
   \]
   \[
   2n \geq 3k
   \]
   \[
   k \leq \frac{2n}{3}
   \]
   Since \( k \) must be an integer, we take the floor function:
   \[
   k \leq \left\lfloor \frac{2n}{3} \right\rfloor
   \]
   Substituting \( n = 2014 \):
   \[
   k \leq \left\lfloor \frac{2 \cdot 2014}{3} \right\rfloor = \left\lfloor \frac{4028}{3} \right\rfloor = 1342
   \]

4. **Constructing an example:**
   To show that \( k = 1342 \) is achievable, we construct a graph with \( 3t + 1 \) vertices where \( t = 671 \). This graph will have degrees \( \{1, 2, \ldots, 2t\} \).

   - Take a vertex \( X \) and connect it to \( 2t \) vertices \( A_1, A_2, \ldots, A_{2t} \).
   - Take another set of \( t \) vertices \( B_1, \ldots, B_t \).
   - Connect \( A_i \) to vertices \( B_1, \ldots, B_{i-1} \) for all \( i = 2, 3, \ldots, t \). This ensures that vertices \( A_1, A_2, \ldots, A_t \) have degrees \( 1, 2, \ldots, t \) respectively.
   - Currently, vertices \( B_1, B_2, \ldots, B_t \) have degrees \( t-1, t-2, \ldots, 0 \) respectively. Connect each \( B_i \) to all vertices \( A_{t+1}, A_{t+2}, \ldots, A_{2t} \) so that their degrees become \( 2t-1, 2t-2, \ldots, t \).

   This construction ensures that the graph has degrees \( \{1, 2, \ldots, 2t\} \) and no 3-cliques.

The final answer is \( \boxed{1342} \)