Example 1 (2001 Irish Mathematical Olympiad) Find the smallest positive integer $a$ such that there exists a positive odd integer $n$ satisfying 2001|
$$
55^{n}+a \cdot 32^{n} \text {. }
$$
Analysis Using the properties of congruence, note that $2001=23 \times 87$, we can find that $a \equiv 1(\bmod 87)$ and $a \equiv-1$ $(\bmod 23)$. Thus, we can obtain the smallest value of $a$ that meets the requirements.
Solution Since $2001=87 \times 23$. From the problem, there exists a positive odd number $n$ such that
$87 \mid 55^{n}+a \cdot 32^{n}$,
and $23 \mid 55^{n}+a \cdot 32^{n}$.
From (1) we know $: 0 \equiv 55^{n}+a \cdot 32^{n} \quad(\bmod 87)$
$$
\begin{array}{l}
\equiv(-32)^{n}+a \cdot 32^{n} \quad(\bmod 87) \\
\equiv 32^{n}(a-1) \quad(\bmod 87) .
\end{array}
$$

Also, $\left(32^{n}, 87\right)=1$, so $a-1 \equiv 0(\bmod 87)$
From (2) we know: $0 \equiv 55^{n}+a \cdot 32^{n}(\bmod 23)$
$$
\equiv 32^{n}+a \cdot 32^{n} \quad(\bmod 23) .
$$

Also, $\left(32^{n}, 23\right)=1$, so $a+1 \equiv 0(\bmod 23)$,
hence $a \equiv-1(\bmod 23)$
From (3), let $a=87 k+1(k \in \mathbf{N})$ and substitute into (4) to get
$$
23|87 k+2 \Rightarrow 23| 18 k+2 \Rightarrow 23|-5 k+25 \Rightarrow 23| k-5
$$

Also, $k \geqslant 0$, so $k \geqslant 5$.
Thus, the smallest value of $a$ is $87 \times 5+1=436$.