Problem - 4324
Find the sum of all $n$ such that $$\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3}+\cdots +\binom{n}{2018} = 0$$
This is a $2018^{th}$ degree polynomial which has $2018$ complex roots. Because all integers $n$ satisfying $1\le n\le 2018$ meet the requirement, therefore the desired answer is $$1+2+\cdots + 2018 = \boxed{2037171}$$