Solution:

Answer: 504510

First, note that none of the numbers will ever be $0$. Let $\star$ denote the replacement operation. For each pair on the board $\left(x_{i}, y_{i}\right)$ define its primary form to be $\left(x_{i}, y_{i}\right)$ and its secondary form to be $\left[x_{i} y_{i}, \frac{x_{i}}{y_{i}}\right]$. Note that the primary form determines the secondary form uniquely and vice versa. In secondary form,

$$\left[a_{1}, b_{1}\right] \star\left[a_{2}, b_{2}\right]=\left(\sqrt{a_{1} b_{1}}, \sqrt{\frac{a_{1}}{b_{1}}}\right) \star\left(\sqrt{a_{2} b_{2}}, \sqrt{\frac{a_{2}}{b_{2}}}\right)=\left(a_{1} b_{2}, \frac{a_{1}}{b_{2}}\right)=\left[a_{1}^{2}, b_{2}^{2}\right].$$

Thus we may replace all pairs on the board by their secondary form and use the above rule for $\star$ instead. From the above rule, we see that if the leftmost number on the board is $x$, then after one minute it will be $x$ or $x^{2}$ depending on whether it was erased in the intervening step, and similarly for the rightmost number. Let $k$ be the number of times the leftmost pair is erased and $n$ be the number of times the rightmost pair is erased. Then the final pair is

$$\left[4022^{2^{k}},\left(\frac{1007}{1006}\right)^{2^{n}}\right]$$

Any step except the last cannot involve both the leftmost and rightmost pair, so $k+n \leq 1005$. Since every pair must be erased at least once, $k, n \geq 1$. Every pair of integers satisfying the above can occur, for example, by making $1005-k-n$ moves involving only the pairs in the middle, then making $k-1$ moves involving the leftmost pair, and finally $n$ moves involving the rightmost pair.

$$\sum_{k=1}^{1004} \sum_{n=1}^{1005-k} 1=\sum_{k=1}^{1004} 1005-k=\sum_{k=1}^{1004} k=\frac{1004 \cdot 1005}{2}=504510$$