Prove that there'll always be three distinct integers $x$, $y$ and $z$ such that $x + y + z = 0$. 
Prove that in $2m + 3$ distinct integers picked from $4m + 3$ integers between $-(2m + 1)$ and $2m + 1$, there'll always be three distinct integers $x$, $y$ and $z$ such that $x + y + z = 0$ ($m \in \mathbb Z^+$).

This problem is adapted from a recent competition. Because of obvious reasons, I decided to alter the problem a little bit. It is certainly not avoided that mistakes will happen. I apologise for that.
Assume that $x < y < z$ and $y = 0$, this case can easily be solved.
But I can't solve the case where $x < y < 0 < z$ or $x < 0 < y < z$. The problem then becomes

$a$ and $m$ are integers such that $0 \ne |a| \le |2m + 1|$. Prove that there'll always two distinct integers $p$ and $q$ such that $-1 < \dfrac{p}{a}, \dfrac{q}{a} \le 0$ and $p + q = -a$.

 A: We try induction on $m$ (with the case $m=0$ being trivial: For the only choice of three out of three ingetgers, we have $-1+0+1=0$).
Let $S$ be the set of $2m+3$ integers picked. 
Assume $0\in S$. Then the other $2m+2$ numbers can have $2m+1$ different absolute values, hence one absolute value occurs twice, say $x\in S$ and $-x\in S$ for some $x\ne 0$. Then $y=0$, $z=-x$ gives us a solution.
Hence we may assume $0\notin S$. 
Assume there are $r$ negatives and $s$ positives in $S$,
$$a_1<a_2<\ldots<a_r<0<b_1<\ldots < b_s.$$
So $r+s=2m+3$ and in particular $r,s\ge 2$. 
If $a_1>-2m$ then $S\setminus\{2m,2m+1\}$ has at least $2(m-1)+3$ integers in the range $-2(m-1)-1,\ldots,2(m-1)+1$ and the claim follows by indcution. Hence we may assume $a_1\le -2m$ and likewise $b_s\ge 2m$.
Now we find $r+s-1=2m+2$ distinct integers that are not "allowed" to be in $S$:
$$\tag1-(a_1+b_1)>-(a_1+b_2)>\ldots>-(a_1+b_s)>-(a_2+b_s)>\ldots>-(a_r+b_s).$$
Note that these all lie in the range $-2m,\ldots,2m$.
So among the $4m+3$ integers in $[-2m-1,2m+1]$, it looks as if we have $2m+3$ integers that are in $S$ and $2m+2$ that are not. As $(2m+3)+(2m+2)>4m+3$, this is impossible. The only way out is that numbers forbidden in $(1)$  are not always distinct from the other summands, i.e., we have at least two exceptional cases where $a_1+b_i+a_1=0$ or $a_1+b_i+b_i=0$ or $a_j+b_s+a_j=0$ or $a_j+b_s+b_s=0$.
As $a_1\le-2m$, the first option would mean $b_i=4m>2m+1$, hence this cannot happen. Likewise $a_i+b_s+b_s=0$ is imposible. So our two exceptions are that $a_1=-2b_i$ and $b_s=-2a_j$ for some $i,j$. It follows that $a_,b_s$ are even, hence $a_1=-2m$, $b_s=2m$. Then $S\setminus\{a_,b_s\}$ is a set of $2(m-1)+3$ integers in the range $-2(m-1)-1,\ldots,2(m-1)+1$ and again we can aply the induction hypothesis. $\square$
