4. For any positive integer $n$, let $D_{n}$ be the set of all positive divisors of $n$, and $f_{i}(n)(i=0,1,2,3)$ 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
$$
f_{0}(m)+f_{1}(m)-f_{2}(m)-f_{3}(m)=2017 .
$$
4. Represent the positive integer $m$ in its standard form:
$$
m=2^{\alpha}\left(\prod_{i=1}^{k} p_{i}^{\beta_{i}}\right)\left(\prod_{j=1}^{l} q_{j}^{\gamma_{j}}\right),
$$

where $\alpha$ can be $0, p_{1}, p_{2}, \cdots, p_{k}$ are all the prime factors of $m$ that are congruent to 1 modulo 4, and $q_{1}, q_{2}, \cdots, q_{l}$ are all the prime factors of $m$ that are congruent to 3 modulo 4. For the cases where $k$ or $l$ is 0, the corresponding product is defined as 1.
$$
\text { Let } P=\prod_{i=1}^{k}\left(1+\beta_{i}\right), Q=\prod_{j=1}^{i}\left(1+\gamma_{j}\right) \text {. }
$$

First, prove the following conclusions:
(1) When $\alpha \geqslant 1$,
$$
\begin{array}{l}
f_{0}(m)=(\alpha-1) P Q, f_{2}(m)=P Q \text {; } \\
\text { (2) } f_{1}(m)+f_{3}(m)=P Q \text {; } \\
\text { (3) } f_{1}(m)=P\left\lceil\frac{Q}{2}\right\rceil .
\end{array}
$$

Consider any positive divisor of $m$
$$
a=2^{\alpha^{\prime}}\left(\prod_{i=1}^{k} p_{i}^{\beta_{i}}\right)\left(\prod_{j=1}^{i} q_{j}^{\gamma_{j}}\right),
$$

where $\alpha^{\prime} \in\{0,1, \cdots, \alpha\}, \beta_{i}^{\prime} \in\left\{0,1, \cdots, \beta_{i}\right\}$ $(1 \leqslant i \leqslant k), \gamma_{j}^{\prime} \in\left\{0,1, \cdots, \gamma_{j}\right\}(1 \leqslant j \leqslant l)$.

When $\alpha \geqslant 1$, $a \in F_{0}(m)$ if and only if $\alpha^{\prime} \geqslant 2$, and $a \in F_{2}(m)$ if and only if $\alpha^{\prime}=1$. The different choices of $\beta_{i}^{\prime}$ and $\gamma_{j}^{\prime}$ (for $1 \leqslant i \leqslant k$ and $1 \leqslant j \leqslant l$) total $P Q$ ways. Therefore, conclusion (1) holds.

Notice that $f_{1}(m)+f_{3}(m)$ is the number of odd divisors of $m$.
Clearly, $f_{1}(m)+f_{3}(m)=P Q$.
Thus, conclusion (2) holds.
The proof of conclusion (3) is similar to the lemma in the third problem of the first year.
Suppose the positive integer $m$ satisfies
$$
f_{0}(m)+f_{1}(m)-f_{2}(m)-f_{3}(m)=2017 .
$$

If $m$ is odd, then $f_{0}(m)=f_{2}(m)=0$.
Using conclusions (2) and (3), we get
$$
\begin{array}{l}
2017=f_{1}(m)-f_{3}(m)=2 f_{1}(m)-P Q \\
=P\left(2\left\lceil\frac{Q}{2}\right\rceil-Q\right) .
\end{array}
$$

Thus, $Q$ must be odd.
$$
\begin{array}{l}
\text { Also, } 2\left\lceil\frac{Q}{2}\right\rceil-Q=1 \text {, then } \\
P=\prod_{i=1}^{k}\left(1+\beta_{i}\right)=2017 . \\
\text { Note that, } 2017 \text { is a prime number, so } \\
k=1, \beta_{1}=2016 \text {. } \\
\text { Therefore, } m=p_{1}^{2016} \prod_{j=1}^{i} q_{j}^{\gamma_{j}} \geqslant 5^{2016} . \\
\text { If } m \text { is even, then by conclusions (1), (2), and (3), we get } \\
2017 \\
=f_{0}(m)-f_{2}(m)-\left(f_{1}(m)+f_{3}(m)\right)+2 f_{1}(m) \\
=(\alpha-1) P Q-P Q-P Q+2 P\left\lceil\frac{Q}{2}\right\rceil \\
=P\left((\alpha-3) Q+2\left\lceil\frac{Q}{2}\right\rceil\right) .
\end{array}
$$

Note that 2017 is a prime number, so
Thus, $m=p_{1}^{2016} \prod_{j=1}^{l} q_{j}^{\gamma_{j}} \geqslant 5^{2016}$.
If $m$ is even, then by conclusions (1), (2), and (3), we get
2017
$$
\begin{array}{l}
=f_{0}(m)-f_{2}(m)-\left(f_{1}(m)+f_{3}(m)\right)+2 f_{1}(m) \\
=(\alpha-1) P Q-P Q-P Q+2 P\left\lceil\frac{Q}{2}\right\rceil \\
=P\left((\alpha-3) Q+2\left\lceil\frac{Q}{2}\right\rceil\right) .
\end{array}
$$

When $P=2017$, we already have $m \geqslant 2 p_{1}^{2016}>5^{2016}$. When $P=1$,
$$
(\alpha-3) Q+2\left\lceil\frac{Q}{2}\right\rceil=2017 \text {. }
$$

Clearly, $(\alpha-3) Q$ is odd, so $Q$ is odd. Simplifying, we get
$$
(\alpha-2) Q=2016=2^{5} \times 3^{2} \times 7,
$$

where $2^{5}$ divides $(\alpha-2)$.
If $\alpha-2>2^{5}$, then $\alpha-2 \geqslant 2^{5} \times 3$, in which case, we already have $m \geqslant 2^{98}$.
If $\alpha-2=2^{5}$, note that,
$$
Q=7 \times 3 \times 3=9 \times 7=21 \times 3=63 \text {. }
$$

We consider the following four cases.
(i) $l=3$, and $1+\gamma_{1}, 1+\gamma_{2}, 1+\gamma_{3}$ are a permutation of $7, 3, 3$. Note that $q_{1}, q_{2}, q_{3}$ are distinct primes congruent to 3 modulo 4, in which case,
$$
\begin{array}{l}
m \geqslant 2^{34} \times 3^{6} \times 7^{2} \times 11^{2} ; \\
\text { (ii) } l=2,\left\{1+\gamma_{1}, 1+\gamma_{2}\right\}=\{9,7\}, \text { in which case, } \\
m \geqslant 2^{34} \times 3^{8} \times 7^{6}>2^{34} \times 3^{6} \times 7^{2} \times 11^{2} ; \\
\text { (iii) } l=2,\left\{1+\gamma_{1}, 1+\gamma_{2}\right\}=\{21,3\}, \text { in which case, } \\
m \geqslant 2^{34} \times 3^{20} \times 7^{2}>2^{34} \times 3^{6} \times 7^{2} \times 11^{2} ; \\
\text { (iv) } l=1,1+\gamma_{1}=63 \text {, in which case, } \\
m \geqslant 2^{34} \times 3^{62}>2^{34} \times 3^{6} \times 7^{2} \times 11^{2} .
\end{array}
$$

From the above discussion, and noting that,
$2^{34} \times 3^{6} \times 7^{2} \times 11^{2}<2^{34} \times 16^{10}<2^{98}$.
Thus, $m \geqslant \min \left\{5^{2016}, 2^{34} \times 3^{6} \times 7^{2} \times 11^{2}\right\}$
$=2^{34} \times 3^{6} \times 7^{2} \times 11^{2}$.
Furthermore, when $m=2^{34} \times 3^{6} \times 7^{2} \times 11^{2}$, $m$ is even, and $\alpha=34, P=1, Q=63$, satisfying equation (1). Therefore, the smallest positive integer $m$ is $2^{34} \times 3^{6} \times 7^{2} \times 11^{2}$.