Find the four digit number?

HINT:

So, we have $$1000a+100a+10b+b=11(100a+b)$$

$\implies 100a+b$ must be divisible by $11\implies 11|(a+b)$ as $100\equiv1\pmod{99}$

As $0\le a,b\le 9, 0\le a+b\le 18\implies a+b=11$

$$\implies11(100a+b)=11(100a+11-a)=11^2(9a+1)$$

So, $9a+1$ must be perfect square

If we let the four-digit number be XXYY, then this number can be expressed as:

$$1000X + 100X + 10Y + Y = 1100X + 11Y = 11(100X + Y) = k^2$$

(since it's a perfect square) In order for this to be true, $100X + Y$ must be the product of $11$ and a perfect square, and looks like $X0Y$. So now our question is "which product of $11$ and a perfect square looks like $X0Y$?" We can test them: $$11 \cdot 16 = 176\\ 11 \cdot 25 = 275\\ 11 \cdot 36 = 396\\ 11 \cdot 49 = 593\\ 11 \cdot 64 = 704\\ 11 \cdot 81 = 891$$ The only one that fits the bill is $704$. This means there is only one four-digit number that works, and it's $7744$. Enjoy!

I recommend programming when numbers are so low.

Here is a Python solution:

>>> list(filter(lambda n: str(n**2)[0] == str(n**2)[1] and \
                          str(n**2)[-1] == str(n**2)[-2], 
                range(int(1000**0.5),int(10000**0.5))
               )
         )
[88]
>>> 88**2
7744

Note that I broke the line for easier readability.

So 7744 is the only solution.

Given $\overline{aabb}=1100a+11b=k^2$, consider mod $4$: $$k\equiv 0,1,2,3 \pmod{4} \\ k^2\equiv 0,1 \pmod{4}\\ 1100a+11b\equiv 3b\equiv 0,1 \pmod{4} \Rightarrow b=0,3,4,7,8 \ \ \ \ \ \ (1)$$ Also, the last digit of $k^2$ can be: $$b=0,1,4,5,6,9 \ \ \ \ \ \ (2)$$
Hence, from $(1)$ and $(2)$: $$b=0 \ \ \text{or} \ \ 4.$$ And: $$k^2=1100a+11b=11(100a+b)=11^2\cdot 9a+11(a+b) \Rightarrow a+b\equiv 0 \pmod{11}.$$ So, $\overline{aabb}=7744$.