Example 17 Find all positive integers not exceeding $10^{3}$, the last two digits of whose square are the same, but not zero.

Find all positive integers not exceeding $10^{3}$, the last two digits of whose square are the same, but not zero.
The last digit of the square of a non-zero integer can only be $1,4,5,6,9$.
The last two digits of the square of an integer cannot be the same odd number: $11,55,99$. Otherwise, for an odd number $n, 4 \mid n^{2}-$ 1, but $10,54,98$ cannot be divisible by 4.
The last two digits of the square of an even number $n$ cannot be 66, otherwise $4 \mid n^{2}$, but $4 \times 10^{2} l+66$.
Thus, only the last two digits being 44 remain. We just need to find such numbers between 1 and 50, because the squares of $m$ and $50 k+m$ have the same last two digits.

Since the last digit of an integer is only 2 or 8 when the last digit of its square is 4. It is easy to find that in the range from 1 to 50, only 12 and 38 have squares with the last two digits being 44. Therefore, numbers of the form $50 k+12$ and $50 k+38$ have the required property.

Within 1000, there are 40 such numbers, just take $k=0,1,2, \cdots, 19$, to get all the required numbers: $12,38,62,88, \cdots, 962,988$.