Is the following correct？This can solve $n$ is lagre enough? Let $|X|=x$ and consider$$T=\left\{Y: Y \subset X, |Y|=\left[\frac{x}{2}\right]\right\},$$and$$S=\left\{\sum_{a\in A}a : A \in T\right\},$$then $|T|=C_{x}^{\left[\frac{x}{2}\right]}$ and $|S|\leq \frac{x}{2}\cdot 2^n$, we notice that $|S|<x\cdot 2^{\frac{3x}{4}}<C_{x}^{\left[\frac{x}{2}\right]}$ when $x>\frac{4}{3}(n-1)$ is large enough by Stirling's formula, then there exist $A,B \in T$, such that $\sum_{a\in A}a=\sum_{b\in B}b$, we select $A'=A \backslash (A\cap B)$ and $B'=B \backslash (A\cap B)$ will be okay. PS: Stirling's formula is: $$n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n, n\rightarrow +\infty$$

@above that seems to be an interesting approach.. Could you however elaborate why $|S| < x \cdot 2^n$? I can't see why that is true.. We prove the following stronger claim. Suppose $S$ is an arbitary set of natural numbers, such that $|S| = n$. Call a number challenging if it can be expressed as the sum of two distinct elements of $S$. Let $X$ be a subset of the challenging numbers of size at least $\frac{4n}{3}$ Observe that this claim solves the original question, by letting $S = \{ 2^1, 2^2, \cdots, 2^{n-1} \}$ Now create a graph on the vertices of $S$. For any element $x = a + b$ where $x \in X, a \in S, b \in S$, joint the vertices $a$ and $b$. Lemma: Their is a walk of even length that doesn't repeat edges whosee first vertex and last vertex are the same By induction on $k$, it can be seen that any graph with $n-1+k$ edges and $n$ vertices has at least $k$ cycles. The base case of $k=1$ is well known, and for the inductive step just pick an arbitary cycle and delete an arbitrary edge from the cycle. This means that our graph has at least $\frac{n}{3} + 1$ cycles, and since each cycle passes through at least 3 vertices, we know that there is a vertex through which $2$ cycles pass. If either of those is an even cycle, our lemma is true, so let us assume that both cycles (let's call them $U, V$) have odd length. Now we take two cases: $U$ and $V$ have no other intersection point; in this case our walk can just be $U+V$ $U$ and $V$ have at least $2$ common intersection points. Since $U,V$ are differect cycles, there has to be a point $k$ that lies in both, but a neighbour of $k$ in $U$ is not a neighbour of $k$ in $V$. Let us say that $r$ is the vertex closest to $k$ in $U$ in that direction such that $r$ lies in $V$. We know that $r$ is not directly adjacent to $k$ in $U$. We claim that this implies the existence of an even cycle. Their are two ways of different parity in $V$ to travel between $k,r$ and there is a way to travel b/w $k,r$ via vertices not in $V$ (via $U$). This gives us an even cycle Thus the lemma is true in both cases. Let us label the walk $w_1, w_2, \cdots w_{2t}$. We let $A \subseteq X$ correspond to the edges $w_1w_2, w_3w_4, \cdots, w_{2t-1}w_{2t}$ and $B \subseteq X$ correspond to the edges $w_2w_3, w_4w_5, \cdots w_{2t}w_1$. Both $A$ and $B$ have $t$ vertices and total sum $= w_1 + w_2 + \cdots + w_{2t}$

So Pranjal you sort the question in Combinatorics section.

How I can learn graph theory? I have not learnt yet.