The $ 2^N$ vertices of the $ N$-dimensional hypercube $ \{0,1\}^N$ are labelled with integers from $ 0$ to $ 2^N \minus{} 1$, by, for $ x \equal{} (x_1,x_2,\ldots ,x_N)\in \{0,1\}^N$,
\[v(x) \equal{} \sum_{k \equal{} 1}^{N}x_k2^{k \minus{} 1}.\]
For which values $ n$, $ 2\leq n \leq 2^n$ can the vertices with labels in the set $ \{v|0\leq v \leq n \minus{} 1\}$ be connected through a Hamiltonian circuit, using edges of the hypercube only?

[i]E. Bazavan & C. Talau[/i]
To solve this problem, we need to determine for which values of \( n \) (where \( 2 \leq n \leq 2^N \)) the vertices with labels in the set \(\{v \mid 0 \leq v \leq n-1\}\) can be connected through a Hamiltonian circuit using edges of the \( N \)-dimensional hypercube.

1. **Understanding the Hypercube and Vertex Labeling**:
   - The \( N \)-dimensional hypercube has \( 2^N \) vertices.
   - Each vertex can be represented as a binary string of length \( N \).
   - The labeling function \( v(x) = \sum_{k=1}^{N} x_k 2^{k-1} \) converts the binary string \( x = (x_1, x_2, \ldots, x_N) \) into an integer.

2. **Hamiltonian Circuit**:
   - A Hamiltonian circuit is a cycle that visits each vertex exactly once and returns to the starting vertex.
   - We need to determine if the vertices labeled from \( 0 \) to \( n-1 \) can form such a cycle.

3. **Odd-Length Cycle Argument**:
   - Each edge in the hypercube connects two vertices whose binary representations differ by exactly one bit.
   - If there were an odd-length cycle, then at least one coordinate would have to change an odd number of times, which is impossible because each coordinate must return to its original value (even number of changes).

4. **Induction and Construction**:
   - We use induction on \( N \) to show that for \( n = 2, 4, 6, 8, \ldots, 2^N \), the vertices can be connected in a Hamiltonian circuit.
   - **Base Case**: For \( N = 1 \), the hypercube is a line segment with vertices \( 0 \) and \( 1 \). The only possible \( n \) values are \( 2 \), and the vertices \( 0 \) and \( 1 \) form a Hamiltonian circuit.
   - **Inductive Step**: Assume that for an \( (N-1) \)-dimensional hypercube, the vertices labeled from \( 0 \) to \( 2^{N-1} - 1 \) can form a Hamiltonian circuit. We need to show that this holds for an \( N \)-dimensional hypercube.
     - Consider the \( N \)-dimensional hypercube as two \( (N-1) \)-dimensional hypercubes connected by edges between corresponding vertices.
     - By the inductive hypothesis, each \( (N-1) \)-dimensional hypercube has a Hamiltonian circuit.
     - We can connect these two circuits by adding an edge between the last vertex of the first circuit and the first vertex of the second circuit, forming a Hamiltonian circuit for the \( N \)-dimensional hypercube.

5. **Conclusion**:
   - The vertices labeled from \( 0 \) to \( n-1 \) can be connected through a Hamiltonian circuit if and only if \( n \) is even.
   - Therefore, the values of \( n \) for which this is possible are \( n = 4, 6, 8, \ldots, 2^N \).

The final answer is \( \boxed{ n = 4, 6, 8, \ldots, 2^N } \).