To solve this problem, we need to determine how many sequences of length $n$ consisting of positive integers are considered "full" according to the defined condition. The condition implies a hierarchical appearance of integers in the sequence, such that if an integer $k$ appears, then $k-1$ must also appear before the last occurrence of $k$.

We can approach the problem inductively:

1. Base Case: For $n = 1$, the only sequence is $[1]$, which trivially satisfies the condition as there are no integers $k \geq 2$.

2. Inductive Step: Assume that for some $n$, all sequences of positive integers of length $n$ are full. Now consider sequences of length $n+1$.

To form a full sequence of length $n+1$, consider placing the number $n+1$ in the sequence. According to the condition, for any occurrence of $n+1$, an $n$ must appear before the last occurrence of $n+1$. The rest of the sequence before placing $n+1$ can be any full sequence of length $n$.

We can insert $n+1$ at any position in the sequence of length $n$, resulting in $(n+1)!$ permutations of sequences.

Thus, each choice of ordering for the integers from $1$ through $n$ is independent in a full sequence, therefore we have $n!$ full sequences for any positive integer $n$.

Hence, the number of full sequences of length $n$ is:
$$\boxed{n!}$$
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