Example 3 Write a two-digit prime number after another two-digit prime number that is not equal to it, to form a four-digit number. It is known that this four-digit number can be divided by half the sum of these two prime numbers. Find all such four-digit numbers.
(1992, Yingchun Cup Mathematics Competition)
Solution: Let the two two-digit prime numbers be $x$ and $y$ $(x \neq y)$. Then the four-digit number obtained by writing $y$ after $x$ is $100x + y$. According to the problem, there exists an integer $k$ such that
$$
100x + y = k \cdot \frac{x + y}{2}.
$$

Therefore, $(x + y) \mid (200x + 2y)$, which means
$$
(x + y) \mid 198x.
$$

Since $x$ and $y$ are distinct primes, we have
$$
(x + y, x) = (y, x) = 1.
$$

Thus, $(x + y) \mid 198$.
Given that $11 \leqslant x \leqslant 97$, $11 \leqslant y \leqslant 97$, $x$ and $y$ are odd, and $x \neq y$, we have $22 < x + y < 194$, and $x + y$ is even.
Since $198 = 2 \times 9 \times 11$, it follows that $x + y = 66$.
Decomposing 66 into the sum of two distinct two-digit primes, we get
$$
66 = 13 + 53 = 19 + 47 = 23 + 43 = 29 + 37.
$$

Therefore, the required four-digit numbers are
$$
\begin{array}{l}
1353, 5313, 1947, 4719, 2343, \\
4323, 2937, 3729.
\end{array}
$$