19.1 .5 * Find a four-digit square number, where the first two digits are the same, and the last two digits are also the same.
Let the required four-digit number be $x=\overline{a a b b}$, then
$$
x=1000 a+100 a+10 b+b=11(100 a+b)=m^{2}, m \in \mathbf{N},
$$

Thus, $11\left|m, 11^{2}\right| x$. Therefore, $100 a+b=99 a+(a+b)$ is divisible by 11, which means $11 \mid a+b$, and $0<a+b \leqslant 18$, so $a+b=11$. Hence, $x=11^{2}(9 a+1)$, which implies $9 a+1$ is a perfect square. Testing $a=1,2, \cdots 9$, we find that only when $a=7$, $9 a+1$ is a perfect square. Therefore, the required four-digit number is $7744=88^{2}$.