4. 23.

Let $\overline{2 a 1 b 9}=m$.
By Fermat's Little Theorem, we know
$$
\begin{array}{l}
m^{12} \equiv 1(\bmod 13) \\
\Rightarrow m^{2019}=\left(m^{12}\right)^{168} m^{3} \equiv m^{3}(\bmod 13) .
\end{array}
$$

Substituting the complete residue system modulo 13, we get
$$
m \equiv 1,3,9(\bmod 13) \text {. }
$$

Note that, $10^{4} 、 10^{3} 、 10^{2} 、 10^{1} 、 10^{0}$ when divided by 13 yield remainders of $3 、-1,-4,-3 、 1$ respectively.
Thus, $\overline{2 a 1 b 9} \equiv 6-a-4-3 b+9$
$$
\equiv 11-a-3 b(\bmod 13) \text {. }
$$

We discuss $m$ in different cases.
(1) If $m \equiv 1(\bmod 13)$, then
$$
\begin{array}{l} 
a+3 b \equiv 10(\bmod 13) \\
\Rightarrow a+3 b= 10,23,36 \\
\Rightarrow(a, b)=(1,3),(4,2),(7,1),(2,7), \\
(5,6),(8,5),(9,9),
\end{array}
$$

a total of 7 pairs;
(2) If $m \equiv 3(\bmod 13)$, then
$$
\begin{array}{l} 
a+3 b \equiv 8(\bmod 13) \\
\Rightarrow(a, b)=(2,2),(5,1),(8,0),(0,7), \\
(3,6),(6,5),(9,4),(7,9),
\end{array}
$$

a total of 8 pairs;
(3) If $m \equiv 9(\bmod 13)$, then
$$
\begin{array}{l} 
a+3 b \equiv 2(\bmod 13) \\
\Rightarrow(a, b)=(2,0),(0,5),(3,4),(6,3), \\
(9,2),(1,9),(4,8),(7,6),
\end{array}
$$

a total of 8 pairs.
Therefore, there are 23 solutions that satisfy the problem.
4. Given that $\overline{2 a 1 b 9}$ represents a five-digit number. Ask: How many ordered pairs of digits $(a, b)$ are there such that
$$
\overline{2 a 1 b 9}^{2019} \equiv 1(\bmod 13) ?
$$