Example 1 Find the maximum positive integer $m$, such that an $m \times m$ square can be exactly divided into 7 rectangles with no common interior points and no remainder, and the 14 side lengths of the 7 rectangles are $1,2,3,4,5,6,7,8,9,10,11,12,13,14$.
First, consider the general case. Let $a_{1}, a_{2}, \cdots, a_{2 n}$ be any permutation of $1,2,3, \cdots, 2 n$. We seek the maximum value of $S_{n} = a_{1} a_{2} + a_{2} a_{3} + \cdots + a_{2 n-1} a_{2 n}$.
$S_{1} = 1 \cdot 2, S_{2} \leqslant \max \{1 \cdot 2 + 3 \cdot 4, 1 \cdot 3 + 2 \cdot 4, 1 \cdot 4 + 2 \cdot 3\} = 1 \cdot 2 + 3 \cdot 4$. We conjecture that
$$
S_{n} \leqslant 1 \cdot 2 + 3 \cdot 4 + 5 \cdot 6 + \cdots + (2 n-1) \cdot (2 n).
$$
For $n=1$ and $2$, it is known from the above that (1) holds.
Assume $S_{k} \leqslant 1 \cdot 2 + 3 \cdot 4 + \cdots + (2 k-1) \cdot 2 k$. Then for $n=k+1$,
$$
S_{k+1} = a_{1} \cdot a_{2} + \cdots + a_{2 k-1} \cdot a_{2 k} + a_{2 k+1} \cdot a_{2 k+2}.
$$
(1) If $2 k+1$ and $2 k+2$ appear in the same term of (2), without loss of generality, let $a_{2 k+1} \cdot a_{2 k+2} = (2 k+1) \cdot (2 k+2)$. Then by the induction hypothesis,
$$
S_{k+1} = S_{k} + (2 k+1)(2 k+2) \leqslant 1 \cdot 2 + 3 \cdot 4 + \cdots + (2 k-1)(2 k) + (2 k+1)(2 k+2).
$$
(2) If $2 k+1$ and $2 k+2$ appear in different terms of (2), without loss of generality, let $a_{2 k-1} = 2 k+1, a_{2 k+1} = 2 k+2$. Since
$$
\begin{aligned}
& \left[a_{2 k} \cdot a_{2 k+2} + (2 k+1)(2 k+2)\right] - \left[(2 k+1) a_{2 k} + (2 k+2) a_{2 k+2}\right] \\
= & \left[(2 k+1) - a_{2 k+2}\right] \left[(2 k+2) - a_{2 k}\right] > 0.
\end{aligned}
$$

Thus, $S_{k+1} < a_{1} \cdot a_{2} + \cdots + a_{2 k-3} \cdot a_{2 k-2} + a_{2 k} \cdot a_{2 k+2} + (2 k+1)(2 k+2)$
$$
\leqslant 1 \cdot 2 + 3 \cdot 4 + \cdots + (2 k-1)(2 k) + (2 k+1)(2 k+2), \text{ so (1) is proved. }
$$

From the above, if an $m \times m$ square can be exactly divided into $n$ rectangles with no common interior points, and if the $2 n$ side lengths are $1,2,3, \cdots, 2 n-1,2 n$, then the area of the square
$$
m^{2} \leqslant 1 \cdot 2 + 3 \cdot 4 + \cdots + (2 n-1)(2 n),
$$

so $m \leqslant \sqrt{1 \cdot 2 + 3 \cdot 4 + \cdots + (2 n-1)(2 n)}$.
In particular, when $n=7$, we have
$$
m \leqslant \sqrt{1 \cdot 2 + 3 \cdot 4 + \cdots + 13 \cdot 14} = \sqrt{504} < 23,
$$

so $m \leqslant 22$.
Note that $22 = 14 + 8 = 13 + 9 = 12 + 10 = 11 + 6 + 5$. The figures 4-12 and 4-13 below show that $m=22$ is possible.
In conclusion, the maximum value of $m$ is 22.