Somewhat a long comment (not a complete answer).
The first thing that I would suggest is that you should work through some examples and see what happens. I'll help with some of the initial steps.
To think about what you're erasing, let $n$ be a number that has been erased. Then, for each pair of factors $ab$ so that $ab=n$, compute $a+b=m$. Next, you find all other ways to write $m$ as a sum $m=c+d$ and then delete each $cd$ found in this way. Then, you continue until there isn't anything more to delete (or you see the pattern).
In your set-up, you start with $3$ and $5$ deleted.
Deleted: $3$ and $5$.
- Using the deleted $3$, you know that $3$ factors as $3=1\cdot 3$. Therefore, $a+b=1+3=4$. The only sums equal to $4$ are $1+3$ and $2+2$. Taking $c=2$ and $d=2$ gives $cd=4$, so $4$ is deleted. Since we've considered all pairs of factors of $3$, there is nothing more that $3$ can tell us.
Deleted: $3$, $4$, and $5$.
- Using the deleted $4$, you know that $4$ factors as $4=1\cdot 4$ or $2\cdot 2$. In the first case, $1+4=5$, and $5$ can be written as $5=1+4$ or $5=2+3$. The first case leads to $1\cdot 4=4$, which is already deleted, but $2\cdot 3=6$, which can now be deleted. On the other hand, using $2\cdot 2=4$, we have two ways to write $4$ as a sum, $4=1+3$ or $4=2+2$. In either case, the product is $1\cdot 3=3$ or $2\cdot 2=4$, both of which have already been deleted. Therefore, there is nothing more that $4$ can tell us.
Deleted: $3$, $4$, $5$, and $6$.
- Using the deleted $5$, you know that $5$ factors as $1\cdot 5=5$. Therefore, the sum of these two is $1+5=6$. There are three ways to write $6$ as a sum, $1+5=6$, $2+4=6$, and $3+3=6$. The first pair has product $1\cdot 5=5$, which has already been deleted. The second pair has product $2\cdot 4=8$, which can now be deleted. The third pair has product $3\cdot 3=9$, which can also be deleted.
Deleted: $3$, $4$, $5$, $6$, $8$, and $9$.
- Using the deleted $6$, you know that $6$ factors as $1\cdot 6=6$ and $2\cdot 3=6$. Therefore, the corresponding sums are $1+6=7$ or $2+3=5$. There are three sums to get $7$, $1+6=7$, $2+5=7$, and $3+4=7$. There are two sums to get $5$, $1+4=5$ and $2+3=5$. These, combined allow one to delete $6$, $10$, $12$, $4$, and 6$.
Deleted: $3$, $4$, $5$, $6$, $8$, $9$, $10$, and $12$.
Continue this until you find a pattern.