Example 2 Try to find all positive integers that the sum of the digits of a perfect square can take.

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Example 2 Try to find all positive integers that the sum of the digits of a perfect square can take.

将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Example 2 Try to find all positive integers that the sum of the digits of a perfect square can take.
First, the remainder of a square number divided by 9 is $0, 1, 4, 7$, so the sum of the digits of a square number divided by 9 leaves a remainder of $0, 1, 4, 7$.

Second, since
$$
\underbrace{9 \cdots 9^{2}}_{k \uparrow}=\underbrace{9 \cdots}_{k-1 \uparrow} 98 \underbrace{0 \cdots 01}_{k-1 \uparrow}
$$

the sum of the digits is $9k$,
the sum of the digits of the square of 1 is 1,
the sum of the digits of the square of 2 is 4,
$$
\underbrace{3 \cdots 35^{2}}_{k \uparrow}=\underbrace{1 \cdots 1}_{k \uparrow} \underbrace{2 \cdots 2}_{k+1 \uparrow} 5 \text{, the sum of the digits is } 3k+7 \text{, }
$$

thus, the sum of the digits of a perfect square can be any positive integer that leaves a remainder of $0, 1, 4, 7$ when divided by 9.