Example 4 There are 68 pairs of non-zero integers on the blackboard. For a positive integer $k$, at most one of the pairs $(k, k)$ and $(-k, -k)$ appears on the blackboard. A student erases some of these 136 numbers so that the sum of any two erased numbers is not 0. It is stipulated that if at least one number from a pair among the 68 pairs is erased, the student scores one point. Find the maximum score the student can achieve.
Note: These 68 pairs can include some identical pairs. ${ }^{[3]}$
(2010, USA Mathematical Olympiad)
Given that $(j, j)$ and $(-j, -j)$ can appear at most as one pair, we can assume that if $(j, j)$ appears, then $j > 0$ (otherwise, replace $j$ with $-j$). For a positive integer $k$, all $k$ or $-k$ can be deleted from the blackboard, but not both. For each $k > 0$, delete $k$ with probability $p$ and $-k$ with probability $1-p$.

Thus, the probability of scoring for each pair is at least $\min \left\{p, 1-p^{2}\right\}$. Therefore, the expected score for 68 pairs is $68 \min \left\{p, 1-p^{2}\right\}$.

Solving the equation $p = 1 - p^{2}$, we get $p = \frac{\sqrt{5} - 1}{2}$.

Therefore, the expected total score is at least
\[ 68 p = 68 \times \frac{\sqrt{5} - 1}{2} > 42. \]
Thus, there must exist a way to achieve a score of 43.

Finally, we provide an example to show that 44 points may not be feasible.
For $1 \leqslant i \leqslant 8$, each pair $(i, i)$ appears 5 times; for $1 \leqslant i \neq j \leqslant 8$, each pair $(-i, -j)$ appears 1 time: a total of $40 + 28 = 68$ pairs.

Suppose $k$ pairs of numbers from 1 to 8 are deleted. Then the first type of pairs can score $5k$ points; and among the second type of pairs, at least $\mathrm{C}_{k}^{2}$ pairs cannot be taken, so the score is $28 - \mathrm{C}_{k}^{2}$. The total score is $5k + 28 - \mathrm{C}_{k}^{2}$, and by completing the square, the maximum value is 43 points.