Problem 6. (8 points) In one move on the board, either two ones are written, or two identical numbers $n$ already written are erased and $n+1$ and $n-1$ are written in their place. What is the minimum number of moves required to obtain the number 2005 on the board? (Initially, nothing is written on the board.)
Solution: We will prove that the minimum number of moves is $\frac{1}{2}\left(2005^{2}+2003^{2}+2002^{2}+\cdots+2^{2}+1^{2}+1^{2}\right)=1342355520$.

We observe that with each move, the sum of the squares of the recorded numbers increases by 2. This is true both when we write two ones and when we replace $n$ and $n$ with $n-1$ and $n+1$ (since $(n-1)^{2}+(n+1)^{2}=n^{2}+n^{2}+2$). Therefore, the sum of the squares of the recorded numbers is even.

We will prove that if we have obtained the number 2005 after the minimum number of moves, then the numbers $2003, 2002, \ldots, 2, 1$ are also on the board. Let us assume the contrary: let the number $n<2004$ be missing from the board. Every number less than 2005, $n$, must have appeared on the board at some point—otherwise, the numbers $n+1, n+2, \ldots, 2005$ would not have appeared. Let us consider the moment when $n$ was last erased. At this move, the numbers $n+1$ and $n-1$ were recorded. Since the number of moves is minimal, we can assert that either $n+1=2005$, or $n+1$ will be used in the future. In the first case, $n=2004$, and in the second case, when the number $n+1$ is used, the number $n$ will reappear on the board, which is a contradiction.

Since $2005^{2}+2003^{2}+\cdots+2^{2}+1^{2}=2005^{2}+\frac{2003 \cdot 2004 \cdot 4007}{6}=2005^{2}+2003 \cdot 334 \cdot 4007$ is an odd number, and the sum of the squares of the recorded numbers is even, there must be at least one more non-zero number on the board in addition to $2005, 2003, 2002, \ldots, 2, 1$. Since each move increases the sum of the squares of the recorded numbers by 2, the minimum number of moves required for the appearance of 2005 is $\frac{1}{2}\left(2005^{2}+2003^{2}+2002^{2}+\ldots+2^{2}+1^{2}+1^{2}\right)$ (when the non-zero numbers on the board are $2005, 2003, 2002, \ldots, 2, 1, 1$).

By induction, we will prove that for $n=4l+2$ or $4l+3$, the numbers $n, n-2, \ldots, 3, 2, 1$ can appear on the board, and for $n=4l$ or $4l+1$, the numbers $n, n-2, \ldots, 3, 2, 1, 1$ can appear (recorded zeros are not considered).

For $n=1$, the statement is true (1, 1 after the first move).

Assume it is true for $n=k$; we will prove it for $n=k+1$. We will only consider the case $k=4l$, and the other cases are analogous. According to the induction hypothesis, we can obtain $4l, 4l-2, \ldots, 2, 1, 1$. We erase two ones and write 2 and 0, then we erase two 2s and write 3 and 1, and so on, until we erase $4l-2$ and $4l-2$ and write $4l-1$ and $4l-3$. At this point, the numbers on the board are $4l, 4l-1, 4l-2, \ldots, 3, 2, 1$. Now we will write two ones and perform the same actions: 1 and 1 are replaced with 2 and 0; 2 and 2 are replaced with 3 and 1, and so on, until $4l$ and $4l$ are replaced with $4l+1$ and $4l-1$. As a result, the non-zero numbers on the board will be $4l+1, 4l-1, 4l-2, \ldots, 3, 2, 1, 1$.