Suppose the contrary. Firstly we show that $ 0\notin X$. By the Pigeonhole Principle, two of the $ n + 1$ elements of $ X$ must belong to the same subset among the sets $ \{i, - i\}$, $ i = \overline{1,n}$ and their sum will be $ 0$. Now we use induction on $ n$. Base cases $ n = 1,2$ are easy to handle. Suppose we have proved the statement for $ n$. We'll prove it for $ n + 1$. If $ \left|\{n + 1, - n - 1\}\cap X\right|\le1$ then $ |X'| = |X\setminus\{n + 1, - n - 1\}|\ge n + 2$ and by the induction hypothesis, we are done. Suppose now both $ n + 1$ and $ - n - 1$ are in $ X$. Let $ e = |A| = \left|\{x|x\in X,x > 0,x\neq n + 1\}\right|$ and $ f = |B| = \left|\{x|x\in X,x < 0,x\neq - n - 1\}\right|$. Analyze $ 2$ cases: I. $ n$ even. Let $ n = 2k$. $ e + f = n + 1 = 2k + 1$. WLOG, $ e\ge k + 1$. Again, by the Pigeonhole Principle, some $ 2$ elements of $ A$ will belong to the same subset among the subsets $ \{1,2k\},\{2,2k - 1\},\ldots,\{k,k + 1\}$ and their sum will be $ 2k + 1\in X$. II. $ n$ odd. Let $ n = 2k + 1$. $ e + f = n + 1 = 2k + 2$. If one of $ e,f$ is greater than $ k + 1$, say $ e > k + 1$ then consider the sets $ \{1,2k + 1\},\{2,2k\},\ldots,\{k,k + 2\},\{k + 1\}$. Again by Pigeonhole Principle, some $ 2$ elements of $ A$ are in the same set and their sum is $ 2k + 2\in X$. Assume $ e = f = k + 1$. If some of the above $ k + 1$ sets contains $ 2$ elements of $ A$ we're done. Otherwise each of them has exactly $ 1$ element of $ A$. Consequently $ k + 1\in A$. Analogously $ - (k + 1)\in B$. So, $ k + 1, - k - 1,2k + 2\in X$ and $ k + 1 = ( - k - 1) + 2k + 2$. Contradiction.

We will try to find a counter-example. Assume that it's true that no counterexample exists for $ n-1$. We cannot pick $ n$. Proof: If we pick $ n$ then we have $ n+1$ numbers left to pick from the union of the sets $ \lbrace n-1,-1\rbrace \lbrace n-2,-2\rbrace....\lbrace 0,-n\rbrace$. Hence we must pick $ 2$ elements from the same subset and since the difference between the two is $ n$, the condition is satisfied. By a symmetrical argument we can show that we cannot pick $ -n$. Hence we are left to pick $ n+2$ numbers from $ \lbrace -(n-1),...n-1 \rbrace$ but since by induction this cannot be done with $ n+2-1$ numbers, it cannot be done with $ n+2$ numbers since on the $ (n+2-1)th$ pick we will fail. Base case $ n=1$: we must pick $ 3$ numbers from $ \lbrace -1,0,1 \rbrace$, hence we pick the whole set and $ 1+(-1)=0$