$2 \cdot 63$ Insert "+" or "-" between $1^{2}, 2^{2}, 3^{2}, \cdots, 1989^{2}$, what is the smallest non-negative number that can be obtained from the resulting sum?
[Solution] Since there are 995 odd numbers in $1^{2}, 2^{2}, 3^{2}, \cdots, 1989^{2}$, the result of the expression obtained by adding "+" and "-" signs will always be odd, so the smallest non-negative value sought is no less than 1.

Since the squares of 4 consecutive natural numbers: $k^{2},(k+1)^{2},(k+2)^{2},(k+3)^{2}$ can be filled with " + " and " - " signs to get the number 4: $k 2-(k+1)^{2}-(k+2)^{2}+(k+3)^{2}=4$, therefore, by adding " + " and " - " signs to the squares of 8 consecutive natural numbers, a sum of 0 or 8 can be obtained.

By grouping the squares of $22^{2}, 23^{2}, \cdots, 1989^{2}$ into sets of 8 consecutive natural numbers and adding "+" and "-" signs, the total sum of each set of 8 numbers can be made 0, and then by the squares of $6^{2}, 7^{2}, \cdots, 21^{2}$, a sum of 16 can be obtained. By the squares of $1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}$, a sum of $-15: -1^{2}+2^{2}-3^{2}+4^{2}-5^{2}=-$ 15. This way, the sum of these 1989 squares is 1. Therefore, 1 is the smallest non-negative value sought.