See here: https://artofproblemsolving.com/community/c4h1739463_brazilian_mathematical_olympiad_n2_7th8th_grade_day_2_p5

Hey, i have a very very good solution for this problem! $Conjecture:$ The sum of the product of the pairs of the numbers written in the board is invariant. $Proof:$ Consider $S$ = the sum of the product of the pairs of numbers written in the board before making the operation with $a$ and $b$, and $S_1 = $ the sum of the product of the pairs of numbers written in the board after making the operation with $a$ and $b$, define x = $S_(a,b)$ thus $S_1$ = $S - x.(a+b) - a.b + (ab/x + a + b).x = S - x.a - x.b - a.b + ab + a.x + b.x = S => S = S_1$, as we wanted! Q.E.D. So, we have that $x.y$ is invariant, so we can only minimize $y$, because this makes $x$ be maximum. Call $n_k$, the number generated after the operation of number $k$, note that $n_k$ is greater than the numbers $a$ and $b$ who were erased from the board in the operation of number $k$, because $n_k = a + b + E$($E$ is a real number) $=>$ $n_k > a$ and $n_k > b$, so everytime the numbers written in the board are greater than the numbers written before, in the start, the smallest number is $1$, so all the numbers in the board are equal or greater than $1 =>$ the minimum value for $y = 1$, to do this, only make the operations with the other numbers, so making $y = 1 => x.y = x =$ the sum of the products of the pairs of the numbers in the board $=$ the sum of the products of the pair before doing any operation $= (1.2 + 1.3 + 1.4 + ... + 1.10 + 2.1 + 2.2 + ... 2.10 + ....... + 10.1 + 10.2 + ... + 10.10)/2 = [(1+2+...+10)^2 - (1^2 + 2^2 + ... + 10^2)]/2 = [55^2 - 10.11.21/6]/2 = [3025 - 385]/2 = 1320.$ Thus $x.y = 1320 =>$ making $y = 1$, we get that the maximum possible value for $x = 1320$. Q.E.D.