Example 1 (2001 Irish Mathematical Olympiad) Find the smallest positive integer $a$ such that there exists a positive odd integer $n$ satisfying $2001 \mid$
$$55^{n}+a \cdot 32^{n}$$
Analysis Using the properties of congruence, noting 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 $2001=87 \times 23$. By 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 have: $0 \equiv 55^{n}+a \cdot 32^{n} \quad(\bmod 87)$
$$\begin{array}{l}
\equiv(-32)^{n}+a \cdot 32^{n} \quad(\bmod 87) \\
\equiv 32^{n}(a-1) \quad(\bmod 87)
\end{array}$$

Since $\left(32^{n}, 87\right)=1$, it follows that $a-1 \equiv 0(\bmod 87)$
From (2), we have: $0 \equiv 55^{n}+a \cdot 32^{n}(\bmod 23)$
$$\equiv 32^{n}+a \cdot 32^{n} \quad(\bmod 23)$$

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

Since $k \geqslant 0$, it follows that $k \geqslant 5$.
Thus, the smallest value of $a$ is $87 \times 5+1=436$.