2. 55 In decimal, find the smallest natural number: its square number starts with 19 and ends with 89
[Solution] The last digit of the required number can only be 3 or 7. We write out the squares of all two-digit numbers ending in 3 and 7, among which only $17^{2}, 33^{2}, 67^{2}, 83^{2}$ end with 89. To make the first two digits of the natural number $x$ be 19, the inequality must hold:
$19 \leqslant x^{2} \cdot 10^{-N}<20$, where $N$ is a natural number. When $N=2k$ and $N=2k+1$, we get the inequalities
$19 \leqslant x^{2} \cdot 10^{-2k}<20$ and $190 \leqslant x^{2} \cdot 10^{-2k}<200$, respectively. Taking the square root, in the first case we get the inequality
$$4.3588989 \cdots \leqslant x \cdot 10^{-k}<4.4721359 \cdots$$

and in the second case, we have
$13.784048 \cdots \leqslant x \cdot 10^{-k}<14.142135 \cdots$.
It is not difficult to see that, due to the change in $k$, there will be infinitely many possibilities. We examine $k=1$ and $k=2$.

If $4.358 \leqslant x \cdot 10^{-1}<4.472$, then $x=44$;
If $4.358 \leqslant x \cdot 10^{-2}<4.472$, then $436 \leqslant x \leqslant 447$;
If $13.784 \cdots \leqslant x \cdot 10^{-1}<14.142 \cdots$, then $138 \leqslant x \leqslant 141$;
If $13.784 \cdots \leqslant x \cdot 10^{-2}<14.142 \cdots$, then $1379 \leqslant x \leqslant 1414$.
When $k=3$, we similarly examine the two possibilities, and so on. Among the $x$ values we obtained, the smallest number whose last two digits are $17, 33, 67, 83$ is $1383\left(1383^{2}=1912689\right)$. Therefore, the required number is 1383.