Example 4 (14th Irish Mathematical Olympiad, 2001) Find the smallest positive integer $a$ such that there exists a positive odd integer $n$ satisfying $2001 \mid\left(55^{n}+a \cdot 32^{n}\right)$.
Since $2001=87 \times 23$, it follows from the problem statement that there exists a positive odd number $n$ such that $87 \mid\left(55^{n}+a \cdot 32^{n}\right)$ and $23 \mid\left(55^{n}+a \cdot 32^{n}\right)$. Therefore,
$$
0 \equiv 55^{n}+a \cdot 32^{n} \equiv(-32)^{n}+a \cdot 32^{n}=32^{n}(a-1)(\bmod 87) \text {, hence } a-1 \equiv 0(\bmod 87) \text {, }
$$

and $0 \equiv 55^{n}+a \cdot 32^{n} \equiv 32^{n}+a \cdot 32^{n}=32^{n}(a+1)(\bmod 23)$, hence $a+1 \equiv 0(\bmod 23)$. The problem then reduces to finding the smallest positive integer $a$ such that $87 \mid(a-1)$ and $23 \mid(a+1)$. For this, we can set $a=87 k+1$. Then, by 23 dividing $a+1=87 k+2=3 \times 23 k-5 k+2$, it follows that 23 divides $5 k-2$. Let $5 k=23 \cdot m+2\left(m \in \mathbf{N}_{+}\right)$, the problem reduces to finding the smallest positive integer $m$ such that $23 \cdot m+2$ is divisible by 5. By trial, the smallest positive integer $m=1$, thus $k=5$, and the smallest positive integer $a=87 \times 5+1=436$.