1. Assume for contradiction: Suppose that the first four terms and the last four terms in the sequence $S$ are not the same. That is, assume $(a_1, a_2, a_3, a_4) \neq (a_{n-3}, a_{n-2}, a_{n-1}, a_n)$.

2. **Maximality of $n$**: Since $n$ is maximal, the sequence $S$ is the longest possible sequence where every two sections of successive 5 terms are different. This implies that any extension of $S$ by adding either a 0 or a 1 at the end would result in a repeated 5-term sequence.

3. Consider the extensions: Consider the sequences formed by appending 0 and 1 to the end of $S$:
$$(a_{n-3}, a_{n-2}, a_{n-1}, a_n, 0) \quad \text{and} \quad (a_{n-3}, a_{n-2}, a_{n-1}, a_n, 1)$$
Since $n$ is maximal, both of these sequences must have already occurred somewhere in $S$.

4. Position of the sequences: Since $(a_1, a_2, a_3, a_4) \neq (a_{n-3}, a_{n-2}, a_{n-1}, a_n)$, there must be some element before these two 5-sequences. By definition, it cannot be $a_{n-4}$.

5. Contradiction: Thus, the element right before both 5-sequences must be $1 - a_{n-4}$. This means that the sequence:
$$(1 - a_{n-4}, a_{n-3}, a_{n-2}, a_{n-1}, a_n)$$
occurs twice in $S$, which contradicts the condition that every two sections of successive 5 terms in $S$ are different.

6. Conclusion: Therefore, our assumption that $(a_1, a_2, a_3, a_4) \neq (a_{n-3}, a_{n-2}, a_{n-1}, a_n)$ must be false. Hence, the first four terms and the last four terms in the sequence $S$ must be the same.

$\blacksquare$