Let $G$ be a connected graph, with degree of all vertices not less then $m \geq 3$, such that there is no path through all vertices of $G$ being in every vertex exactly once. Find the least possible number of vertices of $G.$
1. **Assume the existence of a longest path:**
   Let \( v_1, v_2, \ldots, v_L \) be a longest path in the graph \( G \) such that no vertex is repeated. Suppose for contradiction that \( L \leq 2m \).

2. **Prove the existence of a cycle:**
   We will prove a lemma that there exists a cycle consisting of exactly \( v_1, v_2, \ldots, v_L \).

   - Since \( v_1 \) and \( v_L \) are endpoints of the longest path, they are only connected to vertices within the path \( v_1, v_2, \ldots, v_L \) (otherwise, the path could be extended, contradicting the assumption that it is the longest path).
   - If \( v_1 \) and \( v_L \) are adjacent, then the path \( v_1, v_2, \ldots, v_L, v_1 \) forms a cycle, proving the lemma.
   - Suppose \( v_1 \) and \( v_L \) are not adjacent. By the pigeonhole principle, since each vertex has a degree of at least \( m \geq 3 \), there must exist some \( 1 < i < L-1 \) such that \( v_i \) is connected to \( v_L \) and \( v_{i+1} \) is connected to \( v_1 \). This forms a cycle: \( v_1, v_2, \ldots, v_i, v_L, v_{L-1}, \ldots, v_{i+1}, v_1 \).

3. **Contradiction with the longest path assumption:**
   Since \( G \) is connected and \( L < 2m+1 \), there must exist some vertex \( v_j \) that is connected to some vertex \( u \) not in our cycle. This implies that we can start from \( u \) and walk into the cycle, forming a path with \( L+1 \) vertices, which contradicts the assumption that \( v_1, v_2, \ldots, v_L \) is the longest path.

4. **Conclusion:**
   Therefore, the assumption that \( L \leq 2m \) must be false. Hence, the least possible number of vertices \( n \) in \( G \) such that there is no Hamiltonian path (a path that visits every vertex exactly once) must be at least \( 2m+1 \).

The final answer is \( \boxed{ 2m+1 } \).