Problem:
let $a,b,c,d,e$ be real numbers, now there are $\left(\binom{5}{3}=10\right)$ numbers $$a+b+c,a+b+d,a+b+e,a+c+d,a+c+e,a+d+e,b+c+d,b+c+e,b+d+e,c+d+e$$
Question1:($\textbf{Jérémy Blanc has solve it}$)
Find the least k such that if $k$ out of these $10$ numbers are $0$,then $$a=b=c=d=e=0$$
Question 2:
let $n$ is give postive integer numbers,Assmue that $x_{1},x_{2},\cdots,x_{2n+1}$ be real numbers,now there are $\binom{2n+1}{n+1}$ numbers $$x_{1}+x_{2}+\cdots+x_{n}+x_{n+1},x_{1}+x_{2}+\cdots+x_{n}+x_{n+2},\cdots,x_{n+1}+x_{n+2}+\cdots+x_{2n+1}$$ Find the least $k$ such that if $k$ out of these $\binom{2n+1}{n+1}$ numbers are $0$,then $$x_{1}=x_{2}=\cdots=x_{2n+1}=0$$
Use Jeremy ieda:
if $n=3$,then let $a=-3,b=c=d=e=f=g=1$,then $\binom{6}{3}=20$,so I guess $k\ge 21$?
In general，so I guess $$k\ge \binom{2n}{2}+1$$ such $$x_{1}=n,x_{2}=x_{3}=\cdots=x_{2n+1}=-1$$ I guess is right? Then How prove it?
Answer: $k=7$
$(i)$ If $a=-2$ and $b=c=d=e=1$, then $\binom{4}{2}=6$ of the numbers are zero (all sums involving $a$). Hence, $k\ge 7$.
$(ii)$ In order to show that $k\le 7$, we assume that at least $7$ sums are zero and show that this implies that $a=b=c=d=e=0$.
In your $7$ sums you have $21$ letters, each one belonging to $\{a,b,c,d,e\}$. Hence, there is one letter which appear five times (at least), because $5\times 4=20<21$. We can assume that $a$ appears five times. There are exactly six sums involving $a$, and five of them are zero. Up to permutation, we can assume that $$0=a+b+c=a+b+d=a+b+e=a+c+d=a+c+e$$ (the only sum with $a$ that I did not take is $a+d+e$). We find $b=c=d=e$ and $a=-2b$. Since $7$ sums are zero, there is one of the sum which does not involve $a$, but only $b,c,d,e$, hence is equal to $0=3b$. So we have found $a=b=c=d=e=0$.