A sequence of vertices $v_1,v_2,\ldots,v_k$ in a graph, where $v_i=v_j$ only if $i=j$ and $k$ can be any positive integer, is called a $\textit{cycle}$ if $v_1$ is attached by an edge to $v_2$, $v_2$ to $v_3$, and so on to $v_k$ connected to $v_1$.  Rotations and reflections are distinct: $A,B,C$ is distinct from $A,C,B$ and $B,C,A$.  Supposed a simple graph $G$ has $2013$ vertices and $3013$ edges.  What is the minimal number of cycles possible in $G$?
1. **Understanding the Cyclomatic Number**:
   The cyclomatic number (or circuit rank, or nullity) of a graph \( G \) is given by:
   \[
   r = m - n + c
   \]
   where \( m \) is the number of edges, \( n \) is the number of vertices, and \( c \) is the number of connected components. This number represents the minimum number of cycles in the graph.

2. **Applying the Formula**:
   Given:
   \[
   m = 3013, \quad n = 2013, \quad c = 1
   \]
   Substituting these values into the formula for the cyclomatic number:
   \[
   r = 3013 - 2013 + 1 = 1001
   \]
   This means the graph \( G \) contains at least 1001 cycles.

3. **Constructing the Graph**:
   To achieve exactly 1001 cycles, we need to construct a graph with 2013 vertices and 3013 edges that has precisely 1001 cycles. One way to do this is as follows:
   - Take 2012 edges \( v_1v_k \) for \( 2 \leq k \leq 2013 \).
   - Add 1001 edges \( v_{2k}v_{2k+1} \) for \( 1 \leq k \leq 1001 \).

4. **Counting the Cycles**:
   In this construction, we have precisely 1001 cycles of the form \( v_1v_{2k}v_{2k+1} \) for \( 1 \leq k \leq 1001 \). Each of these cycles is distinct because rotations and reflections are considered distinct.

5. **Verification**:
   - The total number of edges used is \( 2012 + 1001 = 3013 \), which matches the given number of edges.
   - The total number of vertices is 2013, which matches the given number of vertices.
   - The number of cycles is 1001, which is the minimum number of cycles as calculated using the cyclomatic number.

Therefore, the minimal number of cycles possible in the graph \( G \) is 1001.

The final answer is \(\boxed{1001}\)