Example 1 Let $X$ be the set of irreducible proper fractions with a denominator of 800, and $Y$ be the set of irreducible proper fractions with a denominator of 900, and let $A=\{x+y \mid x \in X, y \in Y\}$. Find the smallest denominator of the irreducible fractions in $A$.
【Analysis】This problem is adapted from the 35th Russian Mathematical Olympiad question ${ }^{[1]}$.
Let $x=\frac{a}{800} \in X, y=\frac{b}{900} \in Y$, where,
$$
\begin{array}{l}
1 \leqslant a \leqslant 799, (a, 800)=1, \\
1 \leqslant b \leqslant 899, (b, 900)=1 .
\end{array}
$$

Then $x+y=\frac{9 a+8 b}{7200}$.
Since $7200=2^{5} \times 3^{2} \times 5^{2}$, and both $a$ and $b$ are odd, $9 a+8 b$ is odd, meaning that the factor $2^{5}$ in the denominator remains after simplification.

Because $(b, 3)=1$, we have $(9 a+8 b, 3)=1$, meaning that the factor $3^{2}$ in the denominator remains after simplification.

Thus, when $9 a+8 b$ is a multiple of 25, the denominator of the simplified fraction $\frac{9 a+8 b}{7200}$ reaches its minimum value of 288.
A specific example can be constructed as follows.
$$
\begin{array}{l} 
9 a+8 b \equiv 0(\bmod 25) \\
\Leftrightarrow 8 b \equiv 16 a(\bmod 25) \\
\Leftrightarrow b \equiv 2 a(\bmod 25) .
\end{array}
$$

Taking $a=1, b=77$ works.