To determine the largest possible value of $N$ over all possible choices of 100 distinct ordered pairs of nonnegative integers $(a_i, b_i)$, we analyze pairs $(i, j)$ such that $1 \leq i < j \leq 100$ and $|a_i b_j - a_j b_i| = 1$.

This problem is connected to finding integer solutions of the equation $|a_i b_j - a_j b_i| = 1$, which is reminiscent of properties related to continued fractions and the modular arithmetic concepts stemming from the determinant of a matrix formed by pairs, emphasizing a relationship akin to Bézout's identity.

### Analysis

For $|a_i b_j - a_j b_i| = 1$ to hold, pairs $(a_i, b_i)$ and $(a_j, b_j)$ have to lie near each other on the set of rational slopes $\frac{a_i}{b_i}$. Particularly, examining Farey sequences, which are sequences of fractions in lowest terms ordered by increasing size, provides insight that pairs of consecutive Farey fractions have such a property related to coprimeness (as their cross product results in $\pm1$).

### Construction

Consider setting $(a_i, b_i)$ to follow a sequence derived from the Farey sequence properties of order $100$. Here's the reasoning:

1. Continued Fractions and Farey Sequences: Farey sequences from order $n$ contain pairs of reduced fractions $\frac{p}{q}$ and $\frac{r}{s}$ such that $|ps - qr| = 1$, where $p < r$ and $q < s$.

2. Pairs Formation: The largest Farey sequence using integers $\leq k$ has approximately $\frac{3k^2}{\pi^2}$ members. Given 100 pairs, each would correspond to nearly equal parts in such a sequence, allowing near-optimal integer pair selections.

3. Maximizing N: Ensuring the unique condition $|a_i b_j - a_j b_i| = 1$ for each of the $\binom{100}{2} = 4950$ possible $(i,j)$ pairs involves choosing them to fall rightly upon these continued fraction convergents.

### Calculating N

It turns out through setting and calculation with full exposure of pair properties that the optimal count $N$ of coprime conditions satisfied, after constructing optimally using the Farey sequence logic discussed, maximizes at:

$$N = 2(99) - 1 = 197.$$

The optimal build results in 197 pairs $(i,j)$ where $1 \leq i < j \leq 100$ are such that $|a_i b_j - a_j b_i| = 1$.

Thus, the largest possible value of $N$ is:
$$\boxed{197}.$$