2. Find the smallest positive integer $n$ such that $n^{2}+20 n+19$ is divisible by 2019.
2. $n=2000$.

Solution 1 Note that, $2019=3 \times 673$, and both 3 and 673 are prime numbers, and
$$
n^{2}+20 n+19=(n+19)(n+1) \text {. }
$$

Thus, at least one of $n+19$ and $n+1$ is divisible by 3.

Since $n+19-(n+1)=18$, it follows that both $n+19$ and $n+1$ are divisible by 3.

Furthermore, one of them must also be divisible by 673, indicating that one of them is divisible by 2019.
Therefore, the smallest positive integer $n$ that satisfies the condition is $n=2000$.
Solution 2 Write the expression as
$$
\begin{array}{l}
n^{2}+20 n+19 \equiv 0(\bmod 2019) . \\
\text { Then } n^{2}+20 n+19 \equiv 0(\bmod 3) \\
\Leftrightarrow n^{2}+2 n+1 \equiv 0(\bmod 3) \\
\Leftrightarrow(n+1)^{2} \equiv 0(\bmod 3), \\
n^{2}+20 n+19 \equiv 0(\bmod 673) \\
\Leftrightarrow(n+19)(n+1) \equiv 0(\bmod 673),
\end{array}
$$

where, equation (1) has a unique solution -1 modulo 3, and equation (2) has two solutions -1 and -19 modulo 673.
Note that, both 3 and 673 are prime numbers and
$$
-19 \equiv-1(\bmod 3) \text {. }
$$

Thus, $n \equiv-1$ or $-19(\bmod 2019)$.