1. Base Case:
For $n = 2$, consider the alphabet $\{a, b\}$. The word "aba" has length 3, and it cannot be reduced to a word of the form $abab$ by removing letters. Thus, the maximal length for $n = 2$ is $2 \cdot 2 - 1 = 3$.

2. Induction Hypothesis:
Assume that for an alphabet of $k$ letters, the maximal length of a word satisfying the given conditions is $2k - 1$.

3. Induction Step:
We need to show that for $n = k + 1$, the maximal length of a word is $2(k + 1) - 1 = 2k + 1$.

- Assume that each letter in the word appears at least 3 times. Then, we can extract a subword using only the first $n-1$ letters (i.e., $k$ letters) that satisfies the condition. This subword would have a length of at least $3(k)$, which is greater than $2k + 1$ because $k > 2$. This contradicts the induction hypothesis that the maximal length for $k$ letters is $2k - 1$.

- Therefore, there must be a letter $a$ that appears at most twice. If we disregard $a$ from our word, we are left with a word using only $k$ letters that satisfies the condition. By the induction hypothesis, this word has a length of at most $2k - 1$.

- Adding back the letter $a$ (which appears at most twice), the total length of the word is at most $2k - 1 + 2 = 2k + 1$.

4. Constructive Example:
To show that the length $2n - 1$ is attainable, consider the word $a_1, a_2, \ldots, a_n, a_{n-1}, \ldots, a_1$. This word has length $2n - 1$ and satisfies the condition that no two consecutive letters are the same and it cannot be reduced to a word of the form $abab$ by removing letters.

Conclusion:
By induction, we have shown that the maximal length of a word for an alphabet of $n$ letters, where no two consecutive letters are the same and the word cannot be reduced to a word of the form $abab$ by removing letters, is $2n - 1$.

The final answer is $\boxed{ 2n - 1 }$.