Example 1 Add a “+” or “-” in front of each number in $1,2, \cdots, 1989$. Find the minimum non-negative algebraic sum, and write down the equation.
First, prove that the algebraic sum is odd.
Consider the simplest case: all filled with " + ", at this time,
$$
1+2+\cdots+1989=995 \times 1989
$$

is odd.
For the general case, it is only necessary to adjust some " + " to " - ".

Since $a+b$ and $a-b$ have the same parity, each adjustment does not change the parity of the algebraic sum, i.e., the total sum remains odd.
$$
\begin{array}{l}
\text { and } 1+(2-3-4+5)+(6-7-8+9)+ \\
\cdots+(1986-1987-1988+1989)=1 .
\end{array}
$$

Therefore, this minimum value is 1.