Let $G$ be a directed graph with infinitely many vertices. It is known that for each vertex the outdegree is greater than the indegree. Let $O$ be a fixed vertex of $G$. For an arbitrary positive number $n$, let $V_{n}$ be the number of vertices which can be reached from $O$ passing through at most $n$ edges ( $O$ counts). Find the smallest possible value of $V_{n}$.
1. **Redefine \( V_n \) and Set Up the Problem:**
   - Redefine \( V_n \) to mean the set of vertices that can be reached from \( O \) by passing through at most \( n \) edges.
   - Let \( |V_n| \) denote the number of vertices in \( V_n \).
   - We aim to show that \( |V_n| \geq \left\lfloor \frac{(n+2)^2}{4} \right\rfloor \).

2. **Define Sets \( S_i \):**
   - Let \( S_i \) denote the set of vertices that are exactly \( i \) edges away from \( O \), for \( i = 0, 1, \ldots, n \).
   - Thus, \( S_0 = \{O\} \) and \( V_n = S_0 \cup S_1 \cup \cdots \cup S_n \).

3. **Claim and Proof:**
   - **Claim:** For a vertex \( a \in S_i \), all vertices \( b \) such that \( b \rightarrow a \) satisfy \( b \in S_{i-1} \).
   - **Proof by Contradiction:**
     - Assume there exists a vertex \( a \in S_i \) and a vertex \( b \) such that \( b \rightarrow a \) but \( b \notin S_{i-1} \).
     - Consider the smallest \( i \) for which this fails. We have \( i < n \).
     - We analyze four cases for \( b \):
       1. **Case 1:** \( b \notin V_n \):
          - Consider a vertex \( v_{i+1} \in S_{i+1} \) with \( a \rightarrow v_{i+1} \).
          - Remove edges \( b \rightarrow a \) and \( a \rightarrow v_{i+1} \).
          - Use other vertices to fix any issues.
       2. **Case 2:** \( b \in S_0 \cup S_1 \cup \cdots \cup S_{i-2} \):
          - \( a \) would be at most \( i-1 \) edges away from \( O \), contradiction.
       3. **Case 3:** \( b \in S_j \) for \( i \le j < n \):
          - Consider the longest path starting at \( a \) and contained within \( V_n \).
          - If it forms a cycle, remove all edges of the cycle.
          - Otherwise, let \( v_x \) be the first vertex to enter \( S_{j+1} \).
          - Remove edges \( a \rightarrow v_1, v_1 \rightarrow v_2, \ldots, v_{x-1} \rightarrow v_x \) and add \( a \rightarrow v_x \).
       4. **Case 4:** \( b \in S_n \):
          - Remove \( b \rightarrow a \) and use other vertices to fix \( b \).

4. **Lemma 1:**
   - **Statement:** \( e_i = x_{i-1} + e_{i-1} \) for all \( 1 \le i \le n \).
   - **Proof:**
     - Summing outdegree minus indegree over \( S_{i-1} \), we get \( e_i \ge x_{i-1} + e_{i-1} \).
     - If \( e_i > x_{i-1} + e_{i-1} \), there exists an edge from \( S_{i-1} \) to \( S_i \) that can be removed, contradicting minimality.

5. **Lemma 2:**
   - **Statement:** \( |S_i| \ge \frac{e_i}{|S_{i-1}|} = 1 + \frac{e_{i-1}}{|S_{i-1}|} \) for each \( 1 \le i \le n \).
   - **Proof:**
     - By the Pigeonhole Principle, there exists a vertex \( v \in S_{i-1} \) with outdegree at least \( \frac{e_i}{|S_{i-1}|} \).
     - This implies \( |S_i| \ge \frac{e_i}{|S_{i-1}|} \).
     - Using Lemma 1, we get \( \frac{e_i}{|S_{i-1}|} = 1 + \frac{e_{i-1}}{|S_{i-1}|} \).

6. **Rearranging and Summing:**
   - Rearrange Lemma 2: \( x_{i-1} (x_i - 1) \ge e_{i-1} \).
   - Let \( s_i = x_1 + \cdots + x_i \) for each \( 1 \le i \le n \).
   - From Lemma 1, \( e_i = s_{i-1} \) for each \( 2 \le i \le n \).
   - Using AM-GM inequality and the above, we get:
     \[
     s_i = e_{i-1} + x_i + x_{i-1} \ge e_{i-1} + 1 + 2 \sqrt{(x_i-1)x_{i-1}} = e_{i-1} + 2 \sqrt{e_{i-1}}
     \]
     for each \( 2 \le i \le n \).

7. **Induction:**
   - Verify \( s_1 \ge 1 \) and \( s_2 \ge 3 \).
   - By induction, \( s_n \ge \left\lfloor \frac{(n+2)^2}{4} \right\rfloor - 1 \).
   - Adding back \( x_0 = 1 \), we get \( s_n \ge \left\lfloor \frac{(n+2)^2}{4} \right\rfloor \).

The final answer is \( \boxed{ \left\lfloor \frac{(n+2)^2}{4} \right\rfloor } \).