For $k, n \in \mathbf{Z}_{+}$, let the finite sequence $\left\{a_{k}\right\}$ have $n$ terms, where $a_{k} \leqslant m, a_{k}, m \in \mathbf{Z}_{+}$. Determine the maximum value of the number of terms $n$ such that the sequence $\left\{a_{k}\right\}$ satisfies:
(1) For any term $a_{k}$, if $a_{k-1}$ and $a_{k+1}$ exist, then $a_{k-1} \neq a_{k+1}$;
(2) There do not exist positive integers $i_{1}<i_{2}<i_{3}<i_{4}$ such that $a_{i_{1}}=a_{i_{3}} \neq a_{i_{2}}=a_{i_{4}}$.
The sequence $m, m, m-1, m-1, \cdots, 2,2,1$, $1,2,2, \cdots, m-1, m-1, m, m$ satisfies the requirements, with the number of terms being $n=4 m-2$. Therefore, the maximum value sought is no less than $4 m-2$.

In a sequence that meets the requirements, any three consecutive terms cannot be the same number, and if a term's two neighbors (or one neighbor of the first or last term) are different from the term, then adding a neighbor with the same value after the term results in a new sequence that still meets the requirements, thus increasing the number of terms by 1. Therefore, when seeking the maximum number of terms, it is only necessary to consider sequences where each term is the same as one of its neighbors. For this, it is only necessary to consider sequences where any two adjacent terms are different (because the number of terms in the latter is twice that of the former).

The following is a proof by mathematical induction: If in the sequence $a_{1}, a_{2}$, $\cdots, a_{n}$ any two adjacent terms have different values, and $a_{i} \in$ $\{1,2, \cdots, m\}(i=1,2, \cdots, n)$, and there does not exist a positive integer $i_{1}1$, then by the condition (2) that the sequence $a_{1}, a_{2}, \cdots, a_{n}$ satisfies, we know that any term in the sequence $a_{1}, a_{2}, \cdots, a_{v}$ and any term in the sequence $a_{v+1}, a_{v+2}, \cdots, a_{n-1}$ are different. Hence, $s+t \leqslant k+1$, where $s$ and $t$ are the number of distinct values taken by these two sequences, respectively. Noting that these two sequences both meet the requirements of the proposition, according to the induction hypothesis, we have
$$
\begin{array}{l}
v \leqslant 2 s-1, \text { and } n-1-v \leqslant 2 t-1 \\
\Rightarrow n-1 \leqslant 2(s+t)-2 \\
\Rightarrow n \leqslant 2(s+t)-1 \leqslant 2(k+1)-1,
\end{array}
$$

The proposition also holds.
In summary, the maximum value of the number of terms $n$ is $4 m-2$.