Result of replacing $1$ to $n$ in pairs by the sum?

Question

I have the following problem:

Alice writes the numbers $1, 2, 3, 4, 5, 6, \ldots, n$ on a blackboard. Bob selects two of these numbers, erases both of them, and writes down their sum on the blackboard. For example, if Bob chose the numbers 3 and 4, the blackboard would contain the numbers $1, 2, 5, 6, 7, 7, 8, \ldots, n\quad$ ($3$ and $4$ are removed from the list and $7$ is appended to the list) Bob continues until there is only one number left on the board. What are the possible values of that number in terms of $n$?

I have tried using brute force on this problem with smaller values of $n$, and I seem to obtain that the number left on the blackboard is $\dfrac{n(n + 1)}{2}$. However, I cannot rigorously prove this, and I do not know if there are other possible values for the number. I would appreciate your help with this problem. Thank you!

Answer 1

Notice that the sum of numbers on the black board do not change after each turn.
For example, when consider the numbers $1,2,3,4$. If we erase $3$ and $4$ and replace it with $7$ then the remaining numbers will be $1,2,7$. Notice that the sum of numbers is same for both cases.

So the last number will be the sum of all numbers on the board at the beginning.

Answer 2

As Bob continues till the last number, by then the total sum would be $1+2+...+n= \frac{n(n+1)}{2}$ (the terms maybe added in different order, clearly). This is a standard formula and can be proved by induction. Hope this helps.