Let's determine those five-digit square numbers, where the fourth digit - that is, the tens place value - is half as much as the sum of the other four digits.
Let the sought square number be written with digits as $N^{2}=A B C D E$, where $A>0$ and $A+B+C+E=2 D$, so $D>0$. According to this, on one hand $1002 D=12$, which does not meet the criteria.

Similarly, if we choose $D E=61$, the ending of $N$ can only be one of the numbers 19, 31, 69, and 81, and only the first two can be extended to a number that meets our conditions with the hundreds digit: 219 or 231. With the first, $219^{2}=47961$, which does not meet the criteria, but the second does: $231^{2}=53361$.

If we try $D E=64$, $N$ can end in 08, 58, 42, or 92, so we need to check $108^{2}$, $258^{2}$, and $192^{2}$. Only the first meets our requirements.

Finally, for $D=9$ and $D E=96$, the two-digit ending of $N$ can be 14, 36, 64, or 86, so $N$ itself can be 114, 264, or 186, and the first and last do meet the criteria.

According to all this, the following four square numbers have the required properties: $11664=108^{2}$, $12996=114^{2}$, $34596=186^{2}$, $53361=231^{2}$.

Remarks. 1. The omitted $264^{2}=69696$ in the last step is interesting for its symmetry and because in it $A+B+C+E=3 D$.

2. Our observation about the two-digit square endings can also be stated as follows: if two integers are represented by a pair of points on the number line that are mirror images of each other with respect to a point representing a number of the form $25 k$, then the two-digit ending of the squares of these two numbers is the same.

Similarly, the one-digit ending of the squares of $(5 k \pm n)^{2}$ numbers is the same.