Find the smallest positive integer $a$ that satisfies the following condition: there exists a positive odd number $n$, such that $203^{n} + a \cdot 198^{n}$ is a multiple of 2005.
Solution: Since $2005=401 \times 5$, and
$$
203+198=401,203-198=5 \text {, }
$$

Therefore,
$$
203^{n}+a \cdot 198^{n}=(401-198)^{n}+a \cdot 198^{n}
$$

is a multiple of $401 \times 5$.
Since $n$ is odd, then
$$
-198^{n}+a \cdot 198^{n}=(a-1) \cdot 198^{n}
$$

is a multiple of 401.
Since the greatest common divisor of 198 and 401 is 1, then $a-1$ is a multiple of 401. Also,
$$
203^{n}+a \cdot 198^{n}=(198+5)^{n}+a \cdot 198^{n}
$$

is a multiple of $401 \times 5$, then
$$
198^{n}+a \cdot 198^{n}=(a+1) \cdot 198^{n}
$$

is a multiple of 5.
Since the greatest common divisor of 198 and 5 is 1, then $a+1$ is a multiple of 5.

From the above, $a+1$ is a multiple of 5 and $a-1$ is a multiple of 401, so we can set
$a-1=401 t$ (where $t$ is an integer).
Thus, $a+1=401 t+2$.
Therefore, taking the smallest $t=3, a+1=1205$ is a multiple of 5.
Hence, the smallest $a$ is 1204.