I rember this. I post my solution soon.

Lemma Given a complete graph on $k\ge 3$ vertices whose edges are colored in black and white s.t. every triangle has an even number of black edges (either $0$ or $2$), we can find a complete subgraph on $\left\lceil\frac k2\right\rceil$ vertices all of whose edges are white. Proof It’s easy to see that the partial graph generated by all the vertices and the black edges is a complete bipartite graph. If $A,B$ is its bipartition, the subgraphs induced on $A,B$ are both complete graphs with white edges only, and at least one of $|A|,|B|$ is $\ge\left\lceil\frac k2\right\rceil$, so we’re done. We can now prove the assertion by induction on $n$. Suppose it's true for $n-1$. Construct a graph $G$ on a set $S$ with $k=2^n+1$ vertices, and color the edge $(a,b)$ white if $f(\{a,b\})$ is even and black if it's odd. By applying the lemma, we conclude that $S$ has a subset $\bar S$ with $2^{n-1}+1$ elements s.t. all the pairs of elements in $\bar S$ are mapped onto even numbers. Now divide these numbers by $2$ to get a map from a set with $2^{n-1}+1$ elements to $\{0,1,\ldots,2^{n-2}-1\}$ which still satisfies the condition that for any $x,y,z$, one of $f(\{x,y\}), f(\{y,z\}), f(\{z, x\})$ is the sum of the other two, and we can apply the induction hypothesis.

Also in the condition $2^{n} +1$ can be replaced by $2n +1$ and $2^{n-1} -1$ with $n -1$ . This conculsion is based on another solution- not by induction but directly !

We'll prove the result for the problem using $2n+1$ and $n-1$. We consider a graph with $2n+1$ vertices and label each one with a number; in any triangle, the sum of two labels equals the third. We want to prove that there is a triangle with all labels zero. If there are more than $n$ labels that are zero, then some vertex has at least edges labeled zero, which gives us a triangle with all labels zero. If there are at most $n$ labels that are zero, and no vertex has two edges that are labeled zero, then there is a complete subgraph of at least $n+1$ vertices that contains no zero edges (to see this, for each edge labeled zero, take one of the endpoints and throw away the other, and then take all remaining vertices). It remains to derive a contradiction from the existence of this graph of $n+1$ vertices. Indeed, consider the edge AB with the largest label $k \le n-1$. For each vertex C, the labels on CA and CB add up to $k$. By the pigeonhole principle, there exist vertices X and Y that are not B such that AX and AY have the same label, which we'll call $a$. Then BX and BY also have the same label $b$, and since the label on XY is nonzero, it must be both $2a$ and $2b$. It follows then that the label on XY is $k$, and $a = b = \frac{k}{2}$. Now consider some fifth vertex Z. Out of the labels on ZA, ZB, ZX, and ZY, suppose WLOG that ZA and ZX are the biggest. They are both more than $\frac{k}{2}$, so their difference must equal $\frac{k}{2}$, since that's the label on AX. But then one of the labels ZA and ZX must be at least $k$, which means one of them equals $k$, WLOG the label on ZA. But then the label on ZB must equal $0$, since the label on $AB$ is also $k$. This is our desired contradiction, so the result is proven.

imortal wrote: Also in the condition $ 2^{n} + 1$ can be replaced by $ 2n + 1$ and $ 2^{n - 1} - 1$ with $ n - 1$ . This conculsion is based on another solution- not by induction but directly ! What about the induction ? How we can prove it by induction ?

solved with graph theory but didnt include graph theory in sol