3. (25 points) In the sum $0^{2}+1^{2}+2^{2}+3^{2}+\cdots+$ $2005^{2}$, some of the “+” signs can be changed to “-” signs. If the resulting algebraic sum is $n$, the number $n$ is called “representable”. Which of the first 10 positive integers 1, 2, $3, \cdots, 10$ are representable? Explain your reasoning.
3. In $0^{2}+1^{2}+2^{2}+3^{2}+\cdots+2005^{2}$, there are 1003 (an odd number) odd numbers, so no matter how " + " or " - " signs are added, the algebraic sum must be odd. Therefore, among the first 10 positive integers, the numbers that can be expressed must be among $1,3,5,7,9$.
Below, we show that these 5 numbers can all be expressed.
Notice that for 4 consecutive square numbers
$$
k^{2},(k+1)^{2},(k+2)^{2},(k+3)^{2} \text {, }
$$

we have $k^{2}-(k+1)^{2}-(k+2)^{2}+(k+3)^{2}=4$,
and $-k^{2}+(k+1)^{2}+(k+2)^{2}-(k+3)^{2}=-4$.
Therefore, 8 consecutive square numbers, with appropriate " + " or " - " signs, can have an algebraic sum of 0. Thus, $8 n$ consecutive square numbers, with " + " or " - " signs, can also have an algebraic sum of 0.

Now, take 2000 $(8 \times 250$) consecutive square numbers $6^{2}, 7^{2}, \cdots$, $2005^{2}$, and add " + " or " - " signs to make their algebraic sum 0, then handle the first 5 numbers.
Since $-1^{2}+2^{2}-3^{2}-4^{2}+5^{2}=3$,
$$
1^{2}+2^{2}-3^{2}-4^{2}+5^{2}=5 \text {, }
$$

both 3 and 5 can be expressed.
Take 1992 $\left(8 \times 249\right)$ consecutive square numbers $14^{2}, 15^{2}, \cdots$, $2005^{2}$, and add " + " or " - " signs to make their algebraic sum 0, then handle the first 13 square numbers. Since
$$
\begin{array}{l}
-\left(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}+7^{2}\right)+8^{2}+9^{2}- \\
\left(10^{2}-11^{2}-12^{2}+13^{2}\right)=1, \\
-\left(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}+7^{2}\right)+8^{2}+9^{2}+ \\
\left(10^{2}-11^{2}-12^{2}+13^{2}\right)=9, \\
1^{2}+2^{2}+3^{2}-4^{2}-5^{2}-6^{2}+7^{2}-8^{2}+9^{2}+10^{2}- \\
11^{2}-12^{2}+13^{2}=7,
\end{array}
$$

1, 7, and 9 can also be expressed.
Therefore, $1,3,5,7,9$ can all be expressed.