Example 7: In $1^{2}, 2^{2}, 3^{2}, \cdots, 2005^{2}$, add a “+” or “-” sign before each number to make their algebraic sum the smallest non-negative number, and write out the equation.
Analysis: First, we explore the general pattern.
Solution: Since there are 1003 odd numbers among the 2005 numbers from $1^{2}$ to $2005^{2}$, the sum of these 2005 numbers is odd. Changing some of the “+” signs to “-” signs does not change the parity of the result, i.e., the algebraic sum of these 2005 numbers is odd. Therefore, the smallest non-negative number sought cannot be 0.

Since $n^{2}-(n+1)^{2}-(n+2)^{2}+(n+3)^{2}=4$, by placing “+” signs before the 1st, 4th, 6th, and 7th numbers and “-” signs before the 2nd, 3rd, 5th, and 8th numbers in a sequence of 8 consecutive perfect squares, the algebraic sum of these 8 numbers is 0. Given $2005=8 \times 250+5$, and $-1^{2}+2^{2}-3^{2}+4^{2}-5^{2}=-15$, we can take
$$
\begin{array}{l}
\left(6^{2}-7^{2}-8^{2}+9^{2}\right)+\left(10^{2}-11^{2}-12^{2}+13^{2}\right)+ \\
\left(14^{2}-15^{2}-16^{2}+17^{2}\right)+\left(18^{2}-19^{2}-20^{2}+21^{2}\right)=16 .
\end{array}
$$

From $22^{2}$ to $2005^{2}$, we can divide into 248 groups (each group consisting of 8 numbers), with the algebraic sum of each group of 8 numbers being 0. At this point, the algebraic sum of these 2005 numbers can be 1. That is, the smallest non-negative integer value that the algebraic sum can take is 1.
!Note: Using parity analysis can handle the problem as a whole.