[Solution] Let $n^{2}$ satisfy the requirements of this problem, then
$$
n^{2}=100 a^{2}+b \text {, where } 010 a$,
so $n \geqslant 10 a+1$,
thus $\quad b=n^{2}-100 a^{2} \geqslant 20 a+1$,
if $20 a+141$, then $n^{2}-40^{2} \geqslant 42^{2}-40^{2}>100$, which contradicts $b<100$.
If $n=10 a+1=41$, then $41^{2}=1681$ clearly satisfies the condition.
Therefore, the largest perfect square number sought is 1681.
$2 \cdot 86$ Find the largest perfect square, and it is known that such a perfect square, when the last two digits are subtracted, remains a perfect square (assuming the subtracted digits are not all 0).