Example 3.2.5 Let the sequence $A=a_{1}, a_{2}, \cdots, a_{2005}, a_{i} \in \mathbf{N}^{+}, m(A)$ be the number of $\left(a_{i}, a_{j}, a_{k}\right)(1 \leqslant i<j<k \leqslant 2005)$ satisfying $a_{j}=$ $a_{i}+1, a_{k}=a_{j}+1$. For all possible $A$, find the maximum value of $m(A)$.
Solution: Consider the transformation of $A$ for adjustment
(1) If $a_{i}>a_{i+1}$, then $\left\{a_{1}, a_{2}, \cdots, a_{i}, a_{i+1}, \cdots, a_{2005}\right\} \rightarrow\left\{a_{1}, a_{2}, \cdots\right.$, $\left.a_{i+1}, a_{i}, \cdots, a_{2005}\right\}$ changes $A$ to $A^{\prime}$, it is easy to see that $m(A) \leqslant m\left(A^{\prime}\right)$, so we only need to consider the non-decreasing subsequence $A$.
(2) If $a_{i+1}=a_{i}+1+d$, then $\left\{a_{1}, a_{2}, \cdots, a_{i}, a_{i+1}, \cdots, a_{2005}\right\} \rightarrow\left\{a_{1}+d\right.$, $\left.a_{2}+d, \cdots, a_{i}+d, a_{i+1}, \cdots, a_{2005}\right\}$ changes $A$ to $A^{\prime}$, and we also have $m(A) \leqslant m\left(A^{\prime}\right)$. Therefore, $\max m(A)$ only needs to be considered for the sequence $A$ of the following form
At this point, $m(A)=t_{1} t_{2} t_{3}+t_{2} t_{3} t_{4}+\cdots+t_{s-2} t_{s-1} t_{s}, t_{1}+t_{2}+\cdots+t_{s}=2005$.
(3) If $S>4$, divide 2005 into $S-1$ parts, $t_{2}, t_{3}, t_{1}+t_{4}, t_{5}, \cdots, t_{s}$.
$$
\begin{array}{l}
t_{2} t_{3}\left(t_{1}+t_{4}\right)+t_{3}\left(t_{1}+t_{4}\right) t_{5}+\left(t_{1}+t_{4}\right) t_{5} t_{6}+\sum_{i=6}^{s-1} t_{i-1} t_{i} t_{i+1} \\
>t_{1} t_{2} t_{3}+t_{2} t_{3} t_{4}+t_{3} t_{4} t_{5}+t_{4} t_{5} t_{6}+\sum_{i=6}^{s-1} t_{i-1} t_{i} t_{i+1} .
\end{array}
$$
$m(A)$ increases, so we only need to consider $S=3,4$.
When $S=4$, divide 2005 into 3 parts, $t_{2}, t_{3}, t_{1}+t_{4}$, at this point the value of $m(A)$ remains unchanged, so the maximum value only needs to be considered for $S=3$.
At this point, $m(A)=t_{1} t_{2} t_{3}$, where $t_{1}+t_{2}+t_{3}=2005$.

Therefore,
$$
\begin{array}{c}
\max _{t_{1}+t_{2}+t_{3}=2005} t_{1} t_{2} t_{3}=668^{2} \times 669, \\
\max m(A)=668^{2} \times 669 .
\end{array}
$$

When $A=(\underbrace{a, a, \cdots, a,}_{668} \underbrace{a+1, a+1, \cdots, a+1,}_{668} \underbrace{a+2, a+2, \cdots, a+2}_{669})$, $m(A)$ reaches the maximum value $668^{2} \times 669$.