1. [5] 2021 people are sitting around a circular table. In one move, you may swap the positions of two people sitting next to each other. Determine the minimum number of moves necessary to make each person end up 1000 positions to the left of their original position.
Answer: 1021000
Solution 1: We claim that the answer is $1000 \cdot 1021=1021000$. To see how we can obtain this, label the people around the table $s_{1}, s_{2}, \ldots, s_{2021}$. We then start with $s_{1}$ and swap them 1000 positions to the left, then we take $s_{2}$ and swap them 1000 positions to the left, and so on until we swap $s_{1021} 1000$ positions to the left. We see that after swapping $s_{1} 1000$ positions to the left, we have $s_{1022}, s_{1023}, \ldots, s_{2021}$ still in order followed by $s_{2}, s_{3}, \ldots, s_{1021}, s_{1}$. Further, we can show that after moving $s_{k}$ to the left 1000 times, $s_{k}$ is to the left of $s_{1022}$ with the order $s_{1022}, s_{1023}, \ldots, s_{2021}, s_{k+1}, s_{k+2}, \ldots, s_{1021}, s_{1}, s_{2}, \ldots, s_{k}$ for $1 \leq k \leq 1021$. As we never swap any $s_{k}$ for $1 \leq k \leq 1021$ other than moving it 1000 times to the left, when the operation is done, $s_{1}, s_{2}, \ldots, s_{1021}$ are all shifted 1000 places to the left followed by $s_{1022}, s_{1023}, \ldots, s_{2021}$. Therefore, every person is shifted 1000 to the left and this is done in a total of $1000 \cdot 1021$ moves.
Now, to prove that this is a sufficient lower bound, we note that if a total of $M$ moves are made, then some student will move one seat left $M$ times and one student will move right $M$ times. In order to move every student 1000 seats to the left, we can either have them make at least 1000 total leftward movements or 1021 total rightward movements. This means that there can be at most $\left\lfloor\frac{M}{1000}\right\rfloor$ students who get 1000 seats to the left via a net leftward movement and at most $\left\lfloor\frac{M}{1021}\right\rfloor$ students who get 1000 seats to the left via a net rightward movement. Thus, we see that
$$
\left\lfloor\frac{M}{1000}\right\rfloor+\left\lfloor\frac{M}{1021}\right\rfloor \geq 2021 \Longrightarrow M \geq 1021 \cdot 1000,
$$
as desired.
Solution 2: After getting the same construction as above, we can let the students be labeled $s_{i}$ for $1 \leq i \leq 2021$, and say that in order for each student to end up 1000 spaces to the left, they are moved left $a_{i}$ times and right $b_{i}$ times. Then the condition is trivially equivalent to $a_{i}-b_{i}=1000 \bmod 2021$. Further, as each move will move one person right and one person left, we see that $\sum_{i=1}^{2021} a_{i}$ and $\sum_{i=1}^{2021} b_{i}$ count the same event. Thus,
$$
\sum_{i=1}^{2021} a_{i}=\sum_{i=1}^{2021} b_{i} .
$$

Now, let $a_{i}-b_{i}=2021 c_{i}+1000$ for integer $c_{i}$ and all $1 \leq i \leq 2021$. We see that
$$
0=\sum_{i=1}^{2021} a_{i}-\sum_{i=1}^{2021} b_{i}=\sum_{i=1}^{2021}\left(a_{i}-b_{i}\right)=\sum_{i=1}^{2021}\left(1000+2021 c_{i}\right)=1000 \cdot 2021+2021 \sum_{i=1}^{2021} c_{i} .
$$

Therefore, we find that
$$
\sum_{i=1}^{2021} c_{i}=-1000 .
$$

We then can use the fact that $a_{i}+b_{i} \geq\left|a_{i}-b_{i}\right|=\left|2021 c_{i}+1000\right|$ as both $a_{i}$ and $b_{i}$ are nonnegative. Further, the total number of moves is equal to both $\sum_{i=1}^{2021} a_{i}$ and $\sum_{i=1}^{2021} b_{i}$ so we can then find that
$$
\sum_{i=1}^{2021} \frac{a_{i}+b_{i}}{2} \geq \sum_{i=1}^{2021} \frac{\left|a_{i}-b_{i}\right|}{2}=\sum_{i=1}^{2021} \frac{\left|2021 c_{i}+1000\right|}{2} .
$$

Now, let $\sum_{c_{i} \geq 0} c_{i}=S$ and $\sum_{c_{i}<0} c_{i}=-S-1000$. Then, we find that
$$
\begin{array}{l} 
\sum_{i=1}^{2021} \frac{\left|2021 c_{i}+1000\right|}{2}=\sum_{c_{i} \geq 0} \frac{2021 c_{i}+1000}{2}+\sum_{c_{i}<0} \frac{-2021 c_{i}-1000}{2} \\
= \frac{2021(2 S+1000)}{2}+500\left(-2021+2 \sum_{c_{i} \geq 0} 1\right) .
\end{array}
$$

Combining all of this, we find that
$$
\sum_{i=1}^{2021} \frac{a_{i}+b_{i}}{2} \geq \sum_{i=1}^{2021} \frac{\left|a_{i}-b_{i}\right|}{2} \geq \frac{2021(2 S+1000)}{2}+500\left(-2021+2 \sum_{c_{i} \geq 0} 1\right)=2021 S+1000 \cdot \sum_{c_{i} \geq 0} 1 .
$$

To finish this, we know that if $\sum_{c_{i}<0} c_{i}=-S-1000$, then there are at most $-S-1000$ values of $c_{i}<0$, as the $c_{i}$ with smallest absolute value is -1 . Thus, we find that
$$
2021 S+1000 \cdot \sum_{c_{i} \geq 0} 1 \geq 2021 S+1000(2021-(S+1000))=1021 S+1000 \cdot 1021 .
$$

As $S \geq 0$, we find that this makes the total number of moves, $\sum_{i=1}^{2021} \frac{a_{i}+b_{i}}{2}$ at least $1000 \cdot 1021$. Therefore, we have our desired lower bound and the result follows.