2. Find the smallest positive integer $a$, such that there exists a positive odd integer $n$, satisfying
$$2001 \mid\left(55^{n}+a \cdot 32^{n}\right)$$
2. From $2001=3 \times 23 \times 29$ and the conditions, we have
$$\left\{\begin{array}{ll}
a \equiv 1 & (\bmod 3) \\
a \equiv 1 & (\bmod 29), \\
a \equiv-1 & (\bmod 23)
\end{array}\right.$$

From the first two equations, we can set $a=3 \times 29 \times k+1$, substituting into the last equation gives
$$k \equiv 5(\bmod 23),$$

Thus, we get $a \geqslant 3 \times 29 \times 5+1=436$. When $a=436$, $2001 \mid(55+436 \times 32)$. Therefore, the smallest value is 436.