16.2.17 $\star \star$ Find the smallest decimal natural number $n$, such that its square starts with the digits 19 and ends with the digits 89.
Obviously, the unit digit of $n$ is 3 or 7. Among the two-digit numbers with unit digits of 3 or 7, only 17, $33, 67, 83$ have squares whose last two digits are 89. Therefore, the last two digits of $n$ are $17, 33, 67, 83$. 1989 is not a perfect square, so $n^{2}$ must be at least a five-digit number. If $n^{2}$ is a five-digit number, then $19 \times 10^{3}<n^{2}<20 \times 10^{3}$, which means $137<n<142$, and the last two digits of $n$ do not meet the requirement. If $n^{2}$ is a six-digit number, then $19 \times 10^{4}<n^{2}<20 \times 10^{4}$, which means $435<n<448$, which is also impossible. If $n^{2}$ is a seven-digit number, then $19 \times 10^{5}<n^{2}<20 \times 10^{5}$, which means $1378<n<1415$, so $n=1383$. $n^{2}=1912689$ meets the requirement. Therefore, the smallest natural number sought is 1383.