Strange Perfect Square Number

The given facts mean that if 9 is subtracted from the number it will be divisible by 9, 8, 7, 6 and 5. The lowest common multiple of these is 2520. So the answer must be a multiple of this with 9 added on. The only 4-digit candidates are 2529, 5049 and 7569. Only the last is a square number, being the square of 87.

So 7569.

From the last condition, for a number to be divisible by $5$ , it must end with digit $0$ or $5$ . Thus, the original number must end with $4$ or $9$ .

However, from the second last condition, when subtracted by $3$ , the number will be divisible by $6$ , which is an even number. Therefore, the original number can't end with $4$ . (Otherwise, it will result in $1$ after subtraction.) It must end with $9$ .

Suppose the number is arranged as $\overline{ABC9}$

Now with digit ending with $9$ , when subtracted by $1$ , it will result in ending with $8$ and will be divisible by $8$ . So $\overline{ABC0}$ is also divisible by $8$ ; $\overline{ABC}$ is divisible by $4$ , for there is factor $10$ used.

Similarly, when subtracted by $2$ , it will result in ending with $7$ and will be divisible by $7$ . So $\overline{ABC0}$ is also divisible by $7$ .

Finally, since $\overline{ABC9}$ is divisible by $9$ , the sum of all digits is also a multiple of $9$ . Therefore, $\overline{ABC}$ is also divisible by $9$ .

Taking all three constraints, we will get $\overline{ABC}$ is divisible by $4\times 7\times 9 = 252$ .

That will lead to three possibilities: $252$ ; $504$ ; $756$ .

Of all these outcomes, only $7569 = 3^2 \times 29^2$ is the perfect square number.

As a result, the solution is $\boxed{7569}$ .