What is the largest possible number of edges in a graph on $2n$ nodes, if there exists exactly one way to split its nodes into $n$ pairs so that the nodes from each pair are connected by an edge?

[i]Proposed by Anton Trygub[/i]
1. **Define the problem and notation:**
   We are given a graph \( G \) with \( 2n \) nodes. We need to find the maximum number of edges in \( G \) such that there is exactly one way to split its nodes into \( n \) pairs, with each pair connected by an edge.

2. **Graph partitioning:**
   Let \( A = \{a_1, a_2, \ldots, a_n\} \) and \( B = \{b_1, b_2, \ldots, b_n\} \) be the two sets of nodes such that each \( a_i \) is paired with \( b_i \) by an edge.

3. **Counting edges within sets \( A \) and \( B \):**
   The number of edges within \( A \) and within \( B \) can be at most \( \frac{n(n-1)}{2} \) each, since each set has \( n \) nodes and the maximum number of edges in a complete graph \( K_n \) is \( \frac{n(n-1)}{2} \).

   \[
   |E(G(A))| + |E(G(B))| \leq \frac{n(n-1)}{2} + \frac{n(n-1)}{2} = n(n-1)
   \]

4. **Counting edges between sets \( A \) and \( B \):**
   Consider the edges between \( A \) and \( B \). For each \( a_i \), it is connected to \( b_i \) by definition. Additionally, \( a_i \) can be connected to other \( b_j \) (where \( j \neq i \)), but this must be done in such a way that no other perfect matching is possible.

   If \( a_i \) is connected to \( b_j \) (where \( j \neq i \)), then \( a_j \) cannot be connected to \( b_i \) because it would create another possible matching. Thus, the number of such edges is limited by the same reasoning as within sets \( A \) and \( B \).

   \[
   \text{Edges between } A \text{ and } B \leq \frac{n(n-1)}{2}
   \]

5. **Total number of edges:**
   Summing up all the edges, we get:

   \[
   |E(G)| \leq n(n-1) + n = n^2
   \]

6. **Constructing a graph with \( n^2 \) edges:**
   To achieve this bound, consider the following construction:
   - Let \( a_i \) and \( b_i \) be the vertices with \( i \) running from \( 1 \) to \( n \).
   - Connect \( a_1 \) to all vertices in \( B \).
   - Connect \( a_i \) to all vertices in \( B \) except \( b_1, b_2, \ldots, b_{i-1} \) for \( i \geq 2 \).
   - Draw no edges among the \( b_i \).

   This construction ensures that:
   - The degrees of the vertices are \( 2n-1, 2n-2, \ldots, n, n, n-1, \ldots, 1 \), which sum to \( 2n^2 \).
   - By the handshaking lemma, the total number of edges is \( n^2 \).
   - There is a unique matching \( (a_i, b_i) \) for \( i = 1, 2, \ldots, n \).

   Therefore, the graph has exactly \( n^2 \) edges and satisfies the condition of having a unique perfect matching.

The final answer is \( \boxed{ n^2 } \).