Notice that $\mathrm{lcm}(a,b)=\frac{ab}{\gcd(a,b)}$. Also notice that by the Euclidean Algorithm, $\gcd(a,b)\le b-a$. Hence $\mathrm{lcm}(a,b)\ge \frac{ab}{a-b}$. If we require some least common multiple to be greater than $2024$, it is enough to find $a$ and $b$ such that $\frac{ab}{a-b}>2024$, or, equivalently, $\frac{1}{b}-\frac{1}{a}<\frac{1}{2024}$. Let the numbers be $a_1, a_2, \dots, a_{90}$ with $a_1<a_2<\dots<a_{90}$. We wish to show that for some $i$, $\frac{1}{a_i}-\frac{1}{a_{i+1}}<\frac{1}{2024}$. Fix some $t<90$. Consider $$\sum_{j=t}^{90}\left(\frac{1}{a_j}-\frac{1}{a_{j+1}}\right)=\frac{1}{a_t}-\frac{1}{a_{90}}.$$This implies by Pigeonhole that there is a gap of size at most $$\left(\frac{1}{a_t}-\frac{1}{a_{90}}\right)\frac{1}{90-t}.$$Because $a_t\ge t$ we get that this quantity is at most $\frac{1}{t(90-t)}$. Choosing $t=45$ we get that the quantity is at most $\frac{1}{2025}<\frac{1}{2024}$ and we're done. So, the significance of the number $90$ is that $(90/2)^2>2024$ (just barely!)