Hamilton Avenue has eight houses. On one side of the street are the houses
numbered 1,3,5,7 and directly opposite are houses 2,4,6,8 respectively. An
eccentric postman starts deliveries at house 1 and delivers letters to each of
the houses, finally returning to house 1 for a cup of tea. Throughout the
entire journey he must observe the following rules. The numbers of the houses
delivered to must follow an odd-even-odd-even pattern throughout, each house
except house 1 is visited exactly once (house 1 is visited twice) and the postman
at no time is allowed to cross the road to the house directly opposite. How
many different delivery sequences are possible?
To solve this problem, we need to consider the constraints and rules given for the postman's delivery sequence. Let's break down the problem step-by-step:

1. **Identify the Houses and Constraints:**
   - Houses on one side: \(1, 3, 5, 7\)
   - Houses on the opposite side: \(2, 4, 6, 8\)
   - The postman starts and ends at house \(1\).
   - The delivery sequence must follow an odd-even-odd-even pattern.
   - The postman cannot cross directly to the house opposite.

2. **Determine the Pattern:**
   - The sequence must start with an odd house (since the postman starts at house \(1\)).
   - The sequence must alternate between odd and even houses.
   - Each house (except house \(1\)) is visited exactly once.

3. **Possible Sequences:**
   - The postman starts at house \(1\).
   - The next house must be an even house, but not directly opposite house \(1\) (i.e., not house \(2\)).
   - The possible even houses are \(4, 6, 8\).

4. **Constructing the Sequences:**
   - Let's denote the sequence as \(1, E_1, O_1, E_2, O_2, E_3, O_3, 1\), where \(E_i\) are even houses and \(O_i\) are odd houses.
   - The first even house \(E_1\) can be \(4, 6, 8\).
   - After visiting \(E_1\), the next house \(O_1\) must be an odd house that is not directly opposite \(E_1\).

5. **Counting the Valid Sequences:**
   - For each choice of \(E_1\), we need to count the valid sequences that follow the rules.
   - Let's consider each case:

   **Case 1: \(E_1 = 4\)**
   - Possible odd houses for \(O_1\) are \(3, 5, 7\) (but not directly opposite \(4\)).
   - If \(O_1 = 3\), then \(E_2\) can be \(6, 8\) (but not directly opposite \(3\)).
   - If \(O_1 = 5\), then \(E_2\) can be \(2, 8\) (but not directly opposite \(5\)).
   - If \(O_1 = 7\), then \(E_2\) can be \(2, 6\) (but not directly opposite \(7\)).

   **Case 2: \(E_1 = 6\)**
   - Possible odd houses for \(O_1\) are \(3, 5, 7\) (but not directly opposite \(6\)).
   - If \(O_1 = 3\), then \(E_2\) can be \(4, 8\) (but not directly opposite \(3\)).
   - If \(O_1 = 5\), then \(E_2\) can be \(2, 8\) (but not directly opposite \(5\)).
   - If \(O_1 = 7\), then \(E_2\) can be \(2, 4\) (but not directly opposite \(7\)).

   **Case 3: \(E_1 = 8\)**
   - Possible odd houses for \(O_1\) are \(3, 5, 7\) (but not directly opposite \(8\)).
   - If \(O_1 = 3\), then \(E_2\) can be \(4, 6\) (but not directly opposite \(3\)).
   - If \(O_1 = 5\), then \(E_2\) can be \(2, 6\) (but not directly opposite \(5\)).
   - If \(O_1 = 7\), then \(E_2\) can be \(2, 4\) (but not directly opposite \(7\)).

6. **Counting the Total Sequences:**
   - For each choice of \(E_1\), there are \(3\) choices for \(O_1\).
   - For each choice of \(O_1\), there are \(2\) choices for \(E_2\).
   - For each choice of \(E_2\), there is \(1\) choice for \(O_2\).
   - Therefore, the total number of sequences is \(3 \times 3 \times 2 \times 1 = 18\).

However, we need to consider the final move back to house \(1\). If the postman ends up directly opposite house \(1\), it is an illegal move. This happens in \(6\) of the \(18\) sequences (since there are \(3\) choices for \(E_1\) and \(2\) choices for \(O_1\) that lead to an illegal move).

Thus, the number of valid sequences is \(18 - 6 = 12\).

The final answer is \(\boxed{12}\).