8. We consider all possible sets consisting of 2017 different natural numbers, where in each set no number can be represented as the sum of two other numbers in the set. What is the smallest value that the largest number in such a set can take?

Answer: 4032.
Solution. Let for brevity $n=2017$. Consider the following set of $n$ numbers: $n-1, n, n+1, \ldots$, $2 n-3, 2 n-2$. Since $(n-1) + n = 2 n - 1 > 2 n - 2$, no number in this set can equal the sum of two others, meaning the given set satisfies the problem's condition.

Now, let there be an arbitrary set of $n$ natural numbers that satisfy the problem's condition. We will prove that the largest of these numbers $N$ is not less than $2 n - 2$. Suppose $N \leqslant 2 n - 3$. We will prove that in such a set of numbers, there is a pair of numbers less than $N$ whose sum equals $N$. Let's divide all natural numbers less than $N$ into pairs with a sum equal to $N: (1, N-1); (2, N-2), \ldots$. If there are $n-1$ or more such pairs, then the number of numbers less than $N$ will be no less than $2 n - 2$, which is impossible. Therefore, there are no more than $n-2$ such pairs, and there are $n$ numbers in the set. Thus, there will be at least one pair of numbers in the set whose sum equals $N$.

Thus, in any set of $n$ numbers that satisfy the problem's condition, the largest of these numbers is not less than $2 n - 2$ (and this estimate is achievable). In the case $n=2017$, this number is 4032.

Answer to variant 2: 4030.