, 2024

Solution:

We claim that $(a, b)$ is spicy if and only if both $\operatorname{gcd}(a+1, b-1)=1$ and $\operatorname{gcd}(a-1, b+1)=1$.

To prove the claim, we note that
$$\operatorname{gcd}(a+b, a b+1)=\operatorname{gcd}(a+b, b(-b)+1)=\operatorname{gcd}\left(a+b, b^{2}-1\right)$$
Hence, we have
$$\begin{aligned} \operatorname{gcd}(a+b, a b+1)=1 & \Longleftrightarrow \operatorname{gcd}\left(a+b, b^{2}-1\right)=1 \\ & \Longleftrightarrow \operatorname{gcd}(a+b, b-1)=1 \text{ and } \operatorname{gcd}(a+b, b+1)=1 \\ & \Longleftrightarrow \operatorname{gcd}(a+1, b-1)=1 \text{ and } \operatorname{gcd}(a-1, b+1)=1, \end{aligned}$$
proving the claim.

Thus, $n$ works if and only if all following four conditions hold:
- $\operatorname{gcd}(n+1,98)=1$, or equivalently, $n$ is neither $-1\pmod{2}$ nor $-1\pmod{7}$;
- $\operatorname{gcd}(n-1,100)=1$, or equivalently, $n$ is neither $1\pmod{2}$ nor $1\pmod{5}$;
- $\operatorname{gcd}(n+1,100)=1$, or equivalently, $n$ is neither $-1\pmod{2}$ nor $-1\pmod{5}$; and
- $\operatorname{gcd}(n-1,102)=1$, or equivalently, $n$ is neither $1\pmod{2}$, $1\pmod{3}$, nor $1\pmod{17}$.

Thus, there are $1,2,3,6,17$ possible residues modulo $2,3,5,7$, and $17$, respectively. The residues are uniformly distributed within $\{1,2, \ldots, 2024!\}$. Hence, the answer is
$$\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{5} \cdot \frac{6}{7} \cdot \frac{16}{17} = \frac{96}{595}$$