For any positive integer $n$, let $D_{n}$ be the set of all positive divisors of $n$, and $f_{i}(n)$ be the number of elements in the set
$$
F_{i}(n)=\left\{a \in D_{n} \mid a \equiv i(\bmod 4)\right\}
$$

Find the smallest positive integer $m$ such that
$$
2 f_{1}(m)-f_{2}(m)=2017 \text {. }{ }^{[1]}
$$
(14th China Southeast Mathematical Olympiad)
Solution: Clearly, $f_{2}(m) \neq 0$.
Otherwise, $2 f_{1}(m)=2017$ does not meet the problem's requirements.
Thus, $m$ is an even number.
Assume the standard factorization form of $m$ is
$$
m=2^{\alpha}\left(\prod_{i=1}^{k} p_{i}^{\beta_{i}}\right)\left(\prod_{i=1}^{l} q_{i}^{\gamma_{i}}\right),
$$

where $\alpha \in \mathbf{Z}_{+}, \beta_{i}, \gamma_{i} \in \mathbf{N}, p_{i}, q_{i}$ are distinct primes, and
$$
p_{i} \equiv 1(\bmod 4), q_{i} \equiv 3(\bmod 4).
$$

Considering $p_{i}^{r} \equiv 1(\bmod 4)$,
$$
q_{i}^{2 s-1} \equiv 3(\bmod 4), q_{i}^{2 s} \equiv 1(\bmod 4).
$$

Therefore, we discuss two cases.
(1) If $x \in F_{1}(m)$, then
$$
x=\left(\prod_{i=1}^{k} p_{i}^{x_{i}}\right)\left(\prod_{i=1}^{l} q_{i}^{2 y_{i}}\right),
$$

where $0 \leqslant x_{i} \leqslant \beta_{i}(1 \leqslant i \leqslant k)$,
$$
0 \leqslant 2 y_{i} \leqslant \gamma_{i}(1 \leqslant i \leqslant l).
$$

By the multiplication principle, we have
$$
f_{1}(m)=\left(\prod_{i=1}^{k}\left(1+\beta_{i}\right)\right)\left(\prod_{i=1}^{l}\left(1+\left[\frac{\gamma_{i}}{2}\right]\right)\right)
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$.
(2) If $y \in F_{2}(m)$, then
$$
y=2\left(\prod_{i=1}^{k} p_{i}^{x_{i}}\right)\left(\prod_{i=1}^{l} q_{i}^{y_{i}}\right),
$$

where $0 \leqslant x_{i} \leqslant \beta_{i}(1 \leqslant i \leqslant k)$,
$$
0 \leqslant y_{i} \leqslant \gamma_{i}(1 \leqslant i \leqslant l).
$$

By the multiplication principle, we have
$$
f_{2}(m)=\left(\prod_{i=1}^{k}\left(1+\beta_{i}\right)\right)\left(\prod_{i=1}^{l}\left(1+\gamma_{i}\right)\right)
$$
(To make the positive integer $m$ as small as possible, take $\alpha=1$).
Then $2 f_{1}(m)-f_{2}(m)$
$$
\begin{aligned}
= & \left(\prod_{i=1}^{k}\left(1+\beta_{i}\right)\right) \cdot \\
& \left(2 \prod_{i=1}^{l}\left(1+\left[\frac{\gamma_{i}}{2}\right]\right)-\prod_{i=1}^{l}\left(1+\gamma_{i}\right)\right)
\end{aligned}
$$
$=2017$ (a prime number).
We discuss two cases again.
(i) $\prod_{i=1}^{k}\left(1+\beta_{i}\right)=1$,
$2 \prod_{i=1}^{l}\left(1+\left[\frac{\gamma_{i}}{2}\right]\right)-\prod_{i=1}^{l}\left(1+\gamma_{i}\right)=2017$.
From equation (1), we know $\beta_{i}=0(1 \leqslant i \leqslant k)$.
From equation (2), we know $\gamma_{i}$ are all even numbers, set
$\gamma_{i}=2 \omega_{i}(1 \leqslant i \leqslant l)$.
Thus, $2 \prod_{i=1}^{l}\left(1+\omega_{i}\right)-\prod_{i=1}^{l}\left(1+2 \omega_{i}\right)=2017$.
If $l=1$, then
$2\left(1+\omega_{1}\right)-\left(1+2 \omega_{1}\right)=2017$,
which does not meet the problem's requirements.
If $l \geqslant 2$, then
$$
\begin{array}{l}
2\left(1+\omega_{1}\right)\left(1+\omega_{2}\right)-\left(1+2 \omega_{1}\right)\left(1+2 \omega_{2}\right) \\
=1-2 \omega_{1} \omega_{2}<0.
\end{array}
$$

Thus, $2 \prod_{i=1}^{l}\left(1+\omega_{i}\right)-\prod_{i=1}^{l}\left(1+2 \omega_{i}\right)<0$, which does not meet the problem's requirements.
(ii) $\prod_{i=1}^{k}\left(1+\beta_{i}\right)=2017$,
$2 \prod_{i=1}^{l}\left(1+\left[\frac{\gamma_{i}}{2}\right]\right)-\prod_{i=1}^{l}\left(1+\gamma_{i}\right)=1$.
From equation (1), we know $k=1, \beta_{1}=2016$.
To make the positive integer $m$ as small as possible, take $\gamma_{i}=0$ $(1 \leqslant i \leqslant l)$, then equation (2) also holds.
Thus, $m=2 p_{1}^{2016}$, where $p_{1} \equiv 1(\bmod 4)$.
In summary, the smallest positive integer $m$ is $2 \times 5^{2016}$.