4. Calculate: $\log _{2} \prod_{a=0}^{20182} \prod_{b=0}^{2018}\left(1+\mathrm{e}^{\frac{2 a b k i}{2019}}\right)$.
4. 6725.

Notice,
$$
\begin{array}{l}
\log _{2} \prod_{a=0}^{20182} \prod_{b=0}^{2018}\left(1+\mathrm{e}^{\frac{2 a b \pi j}{2019}}\right) \\
=\log _{2} \prod_{a=1}^{20192} \prod_{b=1}^{2019}\left(1+\mathrm{e}^{\frac{2 a b k x_{j}}{2019}}\right) .
\end{array}
$$

Generalize 2019 to a positive odd number $n$.
Let $(a, n)=d$. Then
$$
a=d a_{1}, n=d n_{1} \text {, and }\left(a_{1}, n_{1}\right)=1 ; a_{1}, 2 a_{1} \text {, }
$$
$\cdots, n_{1} a_{1}$ are a permutation of $1,2, \cdots, n_{1}$ modulo $n_{1}$.

Let $\omega=\mathrm{e}^{\frac{2 x_{1}}{1_{1}}}$, then $\omega^{a_{1}}, \omega^{2 a_{1}}, \cdots, \omega^{n_{1} a_{1}}$ are a permutation of $\omega, \omega^{2}$, $\cdots, \omega^{n_{1}}$.
Lemma If $n$ is a positive odd number, then
$$
\prod_{b=1}^{n}\left(1+\mathrm{e}^{\frac{2 a n k}{n}}\right)=2^{d}(d=(a, n)) \text {. }
$$

Proof Notice,
$$
\begin{array}{l}
\prod_{b=1}^{n_{1}}\left(1+\mathrm{e}^{\frac{2 a n k i}{n}}\right)=\prod_{b=1}^{n_{1}}\left(1+\mathrm{e}^{\frac{2 a_{1} \omega_{n} i}{n_{1}}}\right) \\
=\prod_{b=1}^{n_{1}}\left(1+\omega^{b}\right),
\end{array}
$$

and $\omega, \omega^{2}, \cdots, \omega^{n_{1}}$ are all the roots of $z^{n_{1}}-1=0$.
Thus, $z^{n_{1}}-1=\prod_{b=1}^{n_{1}}\left(z-\omega^{b}\right)$.
Let $z=-1$.
Since $n_{1}$ is odd, so,
$$
\begin{array}{l}
\prod_{b=1}^{n_{1}}\left(1+\omega^{b}\right)=2 \\
\Rightarrow \prod_{b=1}^{n}\left(1+\mathrm{e}^{\frac{2=a n \pi i}{n}}\right) \\
=\prod_{b=1}^{n_{1}}\left(1+\mathrm{e}^{\frac{2 \omega n \pi i}{n}}\right) \prod_{b=n_{1}+1}^{2 n_{1}}\left(1+\mathrm{e}^{\frac{23 n d 1}{n}}\right) \cdots \\
\prod_{b=(d-1) n_{1}+1}^{d n_{1}}\left(1+\mathrm{e}^{\frac{2 d a d-1}{n}}\right) \\
=\left(\prod_{b=1}^{n_{1}}\left(1+e^{\frac{2 a_{1}, k=i}{n_{1}}}\right)\right)^{d}=2^{d} \text {. } \\
\end{array}
$$

The lemma is proved.
According to the lemma, let $u(a)=(a, 2019)$.
$$
\begin{array}{l}
\text { Then } \log _{2} \prod_{a=1}^{20192} \prod_{b=1}^{2019}\left(1+\mathrm{e}^{\frac{2 a b-19}{2019}}\right) \\
=\sum_{a=1}^{2019} \log _{2} \prod_{b=1}^{2019}\left(1+\mathrm{e}^{\frac{2 a b-1}{2019}}\right) \\
=\sum_{a=1}^{2019} u(a) .
\end{array}
$$

For each divisor $d$ of 2019, there are $\varphi\left(\frac{2019}{d}\right)$ integers $a$ in $1 \sim 2019$ such that $(a, 2019)=d$, hence
$$
\sum_{a=1}^{2019} u(a)=\sum_{d \mid 2019} d \varphi\left(\frac{2019}{d}\right) .
$$

Let prime $p=3, q=673$. Then $2019=p q$.
$$
\begin{array}{l}
\text { Thus } \sum_{d \mid p q} d \varphi\left(\frac{p q}{d}\right) \\
=1 \cdot(p-1)(q-1)+p(q-1)+q(p-1)+p q \\
=4 p q-2 p-2 q+1=6725 .
\end{array}
$$

In summary, $\log _{2} \prod_{a=0}^{20182} \prod_{b=0}^{2018}\left(1+\mathrm{e}^{\frac{2 a b E j}{2019}}\right)=6725$.