If I've done it right,the answer should be $2023$ though this might be a little trolling like 24IMO5. $\color{red}\rule{25cm}{1pt}$ $\color{red}\textbf{Strategy of Marvin.}$This part is quite trivial since Marvin only need to ask the $i$th mathematician $(1\le i\le 2023)$ to accomplish his goal. $\color{blue}\rule{25cm}{1pt}$ $\color{blue}\textbf{Strategy of mathematicians.}$We claim that even if all the mathematicians only answer $1$ or $2$,they are still able to confuse Marvin as long as $k\le 2022$. Here's the specific strategy.Assume at some point Marvin asks the $i$th mathematician. $\color{blue}\textbf{Case 1.}$There exist $1\le j\le i+1$th and $i+1\le k\le 2024$th mathematicians haven't been asked before.Then we make the $i$th mathematician answer $2$. $\color{blue}\textbf{Case 2.}$All $1\le j\le i+1$th mathematicians have been asked before.Assume that all $i+1\le k\le i+s$th mathematicians have been asked before,($s$ could be $0$)and we maximize $s$.If $s$ is odd,make the $i$th one answer $2$,and answer $1$ when $s$ is even. By induction it's easy to see that there do exist a construction satisfies the answers in this case. $\color{blue}\textbf{Case 3.}$All $i+1\le k\le 2024$th mathematicians have been asked before.This case is symmetrical to the second case. $\rule{25cm}{1pt}$ When Marvin asked all $2022$ questions,there exist $1\le i<j\le 2024$th mathematicians who haven't been asked.Thus Marvin can only determine the situation of $1\le p\le i-1$th and $j+1\le q\le 2024$th mathematicians ,but since obviously all the $i+1\le k\le j-1$th mathematicians answered $2$,it's ez to construct $2$ possible forms of question distributions that fit all the answers,thus $k\ge 2023$.And we're done.