Example 2 Let $A$ and $B$ be two unequal $2n (n \in \mathbf{N}_{+})$-digit numbers, the first $n$ digits of $A$ equal the last $n$ digits of $B$, and the last $n$ digits of $A$ equal the first $n$ digits of $B$, and $A \mid B$. Find the value of $A+B$.
Let $x$ be the positive integer corresponding to the first $n$ digits of $A$, and $y$ be the positive integer corresponding to the last $n$ digits of $A$. Then
$$
A=10^{n} x+y, B=10^{n} y+x \text {. }
$$

Let $B=k A(2 \leqslant k \leqslant 9)$, i.e.,
$$
10^{n} y+x=k\left(10^{n} x+y\right) \text {. }
$$

Thus, $A+B=(x+y)\left(10^{n}+1\right)$
$$
=A+k A=(k+1)\left(10^{n} x+y\right) \text {. }
$$

If $\left(k+1,10^{n}+1\right)=1$, then
$$
10^{n} x+y=0\left(\bmod 10^{n}+1\right) \text {, }
$$

i.e., $-x+y \equiv 0\left(\bmod 10^{n}+1\right)$.
Noting that $B>A \Rightarrow y>x$, and $y-x1(2 \leqslant d \leqslant 10)$.
Using $d \mid\left(10^{n}+1\right)$, it is easy to see that $d \neq 2,3,4,5,6$, $8,9,10$, i.e., $d$ can only be $7, k=6$.

Clearly, $10^{n}+1 \equiv 0(\bmod 7), n=0,1, \cdots, 5$ is only possible for $n=3$.
Combining with Fermat's Little Theorem, we get $n=3(\bmod 6)$.
Let $m=\frac{10^{n}+1}{7}$. Then
$$
\begin{array}{l}
(x+y)\left(10^{n}+1\right)=7\left(10^{n} x+y\right) \\
\Rightarrow m(x+y)=(7 m-1) x+y \\
\Rightarrow(m-1) y=(6 m-1) x . \\
\text { Since }(m-1,6 m-1)=(m-1,5)=1(m
\end{array}
$$

the unit digit of $m$ can only be 3, so $(m-1,5) \neq 5)$, we get
$$
\left\{\begin{array}{l}
x=(m-1) l, \\
y=(6 m-1) l .
\end{array}\right.
$$

Noting that when $l \geqslant 2$,
$$
y \geqslant 2(6 m-1)>7 m=10^{n}+1 \text {, }
$$

which is impossible. Hence, $l=1$.
In summary, when $n=3(\bmod 6)$,
$$
x=\frac{10^{n}-6}{7}, y=\frac{6 \times 10^{n}-1}{7} \text {. }
$$

Therefore, $A+B=10^{2 n}-1(n \equiv 3(\bmod 6))$.