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This problem takes a good understanding on Modular Arithmetic

Let's assume $n \geq 4$

As $n!$ contains $4$ so $4|n!$

in other words, $n! \equiv 0 \pmod{4}$

Now, we assume $2019+n!=k^2$ where $k$ is a natural number.

So $k^2 \equiv 2019+n! \equiv 3+0 \equiv 3 \pmod{4}$

But $k \equiv 0,1,2,3 \pmod{4}$

So, $k^2 \equiv 0,1 \pmod{4}$

But we see $k^2 \equiv 3 \pmod{4}$

Contradiction!

So, $n<4$

Plugging value of $n$ from $1$ to $3$ we see for $n=3$, $2019+n!$ is a perfect square.

So, $n=3$.