3. Find the smallest natural number $n$ such that the number $n^{2}$ begins with 2019 (i.e., $n^{2}=2019 \ldots$).
3. Since $\sqrt{2019}$ is not a natural number $\left(44^{2}=1936440\) and \(n\) 1420 and \(n<\sqrt{20200} \cdot 10<1430\). Now by examining this interval for \(n\) that satisfies the given conditions, we already find for \(n=1421\) that \(1421^{2}=2019241\).

Therefore, the smallest such natural number \(n\) is \(n=1421\).