Let's determine the largest square number not divisible by 100, such that when the last two digits are removed, we still get a square number.
Let this number be $a^{2}$, and the number formed by the last two digits be $0<c<100$. According to the conditions, $\frac{a^{2}-c}{100}$, and thus $a^{2}-c$ is also a square number, so $a^{2}-c \leq(a-1)^{2}$, which means $a^{2}-(a-1)^{2} \leq c<100$. Expanding this, we get $2a-1<100, a \leq 50$, hence $a^{2} \leq 2500$. Therefore, the number we are looking for can have at most 4 digits, so based on the condition, the first two digits can only be $01,04,09,16,25$. The case $a^{2}=2500$ is not valid because it is divisible by 100. The next possible option is represented by the square numbers starting with 16, that is, $1600=40^{2}$ and $1681=41^{2}$. The latter satisfies all conditions, so the number we are looking for is 1681.