Replace $2024$ by $n$. We claim that the answer is $M=2+\frac{n}{n+1}$. Indeed, taking $a=1-\varepsilon, b=n+1-\varepsilon$ and $c=\frac{n}{ab} \approx \frac{n}{n+1}$ for $\varepsilon>0$ sufficiently small, we see that $M \ge 2+\frac{n}{n+1}$ is necessary. On the other hand, let us prove that the inequality always holds for this value of $M$. Indeed, as long as $\lfloor a\rfloor, \lfloor b\rfloor, \lfloor c\rfloor=\left\lfloor \frac{n}{ab}\right\rfloor$, the function $\{a\}+\{b\}+\left\{\frac{n}{ab}\right\}=a+b+\frac{n}{ab}-C$ is convex in each of $a,b$ so that we may always shift two of the three variables until two of them are close to an integer. If one of them is close to an integer from above, the sum is $\le 2+\varepsilon<M$. Finally, if w.l.o.g. $a,b$ are close to an integer from below, say $a=u-\varepsilon$ and $b=v-\varepsilon$, then $c=\frac{n}{uv}+\varepsilon'$ so that the inequality holds with $M=2+\left\{\frac{n}{uv}\right\} \le 2+\frac{n}{n+1}$ as desired. (Note: The same would work with any number of variables.)