Solution:

The number of copies of $n$ in row $n$ is $\phi(n)$, where $\phi$ is the totient function, or the number of positive integers less than or equal to $n$ that are relatively prime to $n$. To see this, consider the following pattern of fractions, where we start with $\frac{0}{1}$ and $\frac{1}{1}$ and insert $\frac{a+c}{b+d}$ between $\frac{a}{b}$ and $\frac{c}{d}$:

The denominators of these fractions form the pattern of numbers in the problem, and the fractions are in increasing order. Row $n$ contains all the fractions in the $n$th Farey sequence exactly once, i.e. all the fractions with denominator at most $n$ (plus some other fractions). Thus, the number of copies of $n$ in row $n$ equals the number of fractions between 0 and 1 with denominator $n$ in simplest form. Each of these fractions must have numerator less than or equal to $n$ and relatively prime to $n$, so there are $\phi(n)$ such fractions. Thus, there are $\phi(n)$ copies of $n$ in row $n$.

The prime factorization of $2019$ is $3 \cdot 673$, thus in row $2019$ the number of copies of $2019$ is
$$\phi(2019) = 2019\left(1-\frac{1}{3}\right)\left(1-\frac{1}{673}\right) = 2 \cdot 672 = 1244$$