Let $G$ be a connected simple graph. When we add an edge to $G$ (between two unconnected vertices), then using at most $17$ edges we can reach any vertex from any other vertex. Find the maximum number of edges to be used to reach any vertex from any other vertex in the original graph, i.e. in the graph before we add an edge.
1. **Understanding the Problem:**
   We are given a connected simple graph \( G \) and we know that by adding an edge \( e \) between two unconnected vertices, the diameter of the new graph \( G+e \) is at most 17. We need to find the maximum possible diameter of the original graph \( G \).

2. **Definitions and Notations:**
   - The diameter of a graph \( \textrm{diam}(G) \) is the greatest distance between any pair of vertices in \( G \).
   - Let \( e = ab \) be the added edge.
   - Let \( P = a \ldots b \) be the shortest path between \( a \) and \( b \) in \( G \) before adding \( e \).
   - Let \( \ell \) be the length of \( P \), i.e., \( \ell = \textrm{d}_G(a, b) \).

3. **Bounding the Length of \( P \):**
   Since \( P \) is the shortest path between \( a \) and \( b \) in \( G \), we have:
   \[
   \ell \leq \textrm{diam}(G)
   \]
   After adding the edge \( e \), the new graph \( G+e \) has a diameter of at most 17:
   \[
   \textrm{diam}(G+e) \leq 17
   \]

4. **Analyzing the Path \( P \):**
   Let \( c \) be a vertex on \( P \) such that the path from \( a \) to \( c \) is half the length of \( P \):
   \[
   \textrm{d}_G(a, c) = \left\lfloor \frac{\ell}{2} \right\rfloor \quad \text{and} \quad \textrm{d}_G(b, c) = \left\lceil \frac{\ell}{2} \right\rceil
   \]

5. **Distance Analysis:**
   For any two vertices \( x \) and \( y \) in \( G \):
   - If the shortest path from \( x \) to \( y \) in \( G+e \) does not use the edge \( e \), then the distance remains the same as in \( G \).
   - If the shortest path from \( x \) to \( y \) in \( G+e \) uses the edge \( e \), then it must pass through either \( a \) or \( b \).

6. **Bounding the Distances:**
   - Consider the shortest path from \( x \) to \( c \) in \( G \). If this path uses \( e \), it must be of the form \( x \ldots a \ldots c \) or \( x \ldots b \ldots c \).
   - In either case, the distance from \( x \) to \( c \) in \( G \) is at most 17:
     \[
     \textrm{d}_G(x, c) \leq 17
     \]
   - Similarly, for \( y \):
     \[
     \textrm{d}_G(y, c) \leq 17
     \]

7. **Combining the Distances:**
   Using the triangle inequality:
   \[
   \textrm{d}_G(x, y) \leq \textrm{d}_G(x, c) + \textrm{d}_G(y, c) \leq 17 + 17 = 34
   \]

8. **Conclusion:**
   Therefore, the diameter of the original graph \( G \) is at most 34:
   \[
   \textrm{diam}(G) \leq 34
   \]

The final answer is \( \boxed{ \textrm{diam}(G) \leq 34 } \)