A modified problem of IMO (Reuploaded)

If $m \ge n+3$ then $n+1,m-1$ are distinct and $(m-1)(n+1) > mn$ .

Consider a (we can assume increasing) sequence of distinct positive integers that add to $2019$ . From the above, if any two successive integers $a<b$ differ by $3$ or more, then we can replace them by $a+1,b-1$ to get another collection of distinct positive integers that add to $2019$ , but which have a bigger product. Similarly, if we can find two pairs of successive integers in the list, $a < b$ and $c < d$ with $b < c$ , where $b \ge a+2$ and $d \ge c+2$ , then we can replace $a,d$ by $a+1,d-1$ , obtaining another list of distinct positive integers that add to $2019$ with a larger product.

Thus, to obtain the maximum product, the integers must be consecutive, except for perhaps one gap where the numbers differ by $2$ .

If the first element of the list was $1$ , we could remove it, and add $1$ to the final element of the list, obtaining a list with sum $2019$ and a larger product. If the first element $a$ of the list was $5$ or greater, we could replace it with $2,a-2$ and obtain a list with sum $2019$ and a greater product.

Thus, to obtain the maximum product, the numbers must be consecutive, except perhaps for one gap where the numbers differ by $2$ , and the first number must be $2$ , $3$ or $4$ .

There are three ways of doing this:

• the numbers from $2$ to $64$ , excluding $60$ ,
• the numbers from $3$ to $64$ , excluding $58$ ,
• the numbers from $4$ to $64$ , excluding $55$ .

Of these, the first option has the largest product $\frac{64!}{60}$ , Thus $a=64$ and $b=60$ , which makes $a+b=\boxed{124}$ .