How many numbers is qualified?

Let $n^2$ have digit sum 2019; therefore $n^2$ must be divisible by 3. If $n^2$ is divisible by 3, then $n$ is divisible by three. Thus $n^2$ is divisible by 9. But $n^2\equiv 2019 \equiv 3\bmod 9$ , a contradiction.

if $x^2$ is the square number and $S(x^2)$ is the sum of the digits of $x^2$ , then

$x^2 \equiv S(x^2) \equiv 2019 \equiv 3\ (mod \ 9) \implies x^2 \equiv 3 \ (mod \ 9) \ (*)$

we know that $x$ cannot be a multiple of $3$ , otherwise $x^2 \equiv 0 \ mod \ 9$ . Therefore $x^2 \equiv 1 \ (mod \ 3)$ , according to Fermat little theorem, which means

$3 | x^2-1 \implies 3 \nmid x^2-1-2 \implies 3 \nmid x^2-3 \implies 9 \nmid x^2-3 \implies x^2 \not\equiv 3 \ (mod \ 9) (**)$

$(*)$ and $(**)$ make a contradiction that suggests there should be no such squares.