假设 $\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right), \ldots,\left(a_{100}, b_{100}\right)$ 是不同的非负整数有序对。令 $N$ 表示满足 $1 \leq i<j \leq 100$ 和 $\left|a_{i} b_{j}-a_{j} b_{i}\right|=1$ 的整数对 $(i, j)$ 的数量。确定在所有可能选择的 100 个有序对中，$N$ 的最大可能值。
The answer is 197. In general, if 100 is replaced by \( n \geq 2 \) the answer is \( 2n - 3 \). The idea is that if we let \( P_{i} = (a_{i}, b_{i}) \) be a point in the coordinate plane, and let \( O = (0,0) \) then we wish to maximize the number of triangles \( \triangle O P_{i} P_{j} \) which have area \( 1 / 2 \). Call such a triangle good. 

【 Construction of 197 points. It suffices to use the points \( (1,0), (1,1), (2,1), (3,1) \), ..., \( (99,1) \) as shown. Notice that:
- There are 98 good triangles with vertices \( (0,0), (k, 1) \) and \( (k+1,1) \) for \( k = 1, \ldots, 98 \).
- There are 99 good triangles with vertices \( (0,0), (1,0) \) and \( (k, 1) \) for \( k = 1, \ldots, 99 \).
This is a total of \( 98 + 99 = 197 \) triangles. 
![](https://cdn.mathpix.com/cropped/2024_11_19_7a2c3e8ae10363caaf59g-10.jpg?height=326&width=818&top_left_y=1382&top_left_x=625)

For the inductive step, suppose (without loss of generality) that the point \( P = P_{n} = (a, b) \) is the farthest away from the point \( O \) among all points.

Claim - This farthest point \( P = P_{n} \) is part of at most two good triangles.
![](https://cdn.mathpix.com/cropped/2024_11_19_7a2c3e8ae10363caaf59g-11.jpg?height=995&width=1000&top_left_y=245&top_left_x=528)
Since \( \operatorname{gcd}(a, b) = 1 \), see that only two lattice points on this locus actually lie inside the quarter-circle centered at \( O \) with radius \( OP \). Indeed if one of the points is \( (u, v) \) then the others on the line are \( (u \pm a, v \pm b) \) where the signs match. This proves the claim.

This claim allows us to complete the induction by simply deleting \( P_{n} \).