Problem: Find, with proof, the least integer $N$ such that
if any $2016$ elements are removed
from the set $\{1, 2, \dots, N\}$,
one can still find $2016$ distinct numbers
among the remaining elements with sum $N$.

Solution: The answer is
\[ N = 2017 + 2018 + \dots + 4032
  = 1008 \cdot 6049 = 6097392. \]

\paragraph{Proof that $N \ge 6097392$ is necessary.}
To see that $N$ must be at least this large,
consider the situation
when $1$, $2$, \dots, $2016$ are removed.
Among the remaining elements,
any sum of $2016$ elements is certainly
at least $2017 + 2018 + \dots + 4032$.

\paragraph{Proof that $N = 6097392$ does in fact work.}
Consider the $3024$ pairs of numbers
$(1, 6048)$, $(2, 6047)$, \dots, $(3024, 3025)$.
Regardless of which $2016$ elements of $\{1, 2, \dots, N\}$ are deleted,
at least $3024 - 2016 = 1008$ of these pairs have both elements remaining.
Since each pair has sum $6049$, we can take these pairs to be the desired numbers.