See also StackExchange.

**Setup.** Let $n\in\Bbb N$. Let $a_{1,1}, a_{1,2},\dots, a_{1,n}\in\Bbb R$ be a given sequence of real numbers that sum to $0$, i.e. $a_{1,n}=-(a_{1,1}+a_{1,2}+\dots+a_{1,n-1})$. For $i=2,\dots,n$ define
$$a_{i,j}=a_{1,j}+a_{1,j+1}+\dots+a_{1,j+i-1}=\sum_{k=j}^{j+i-1} a_{1,k}\quad(\text{for } j=1,\dots,n-i+1).$$
The "half-matrix" $(a_{i,j})_{i,j}$ can be visualized as follows:
$$
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & \dots & a_{1,n-2} & a_{1,n-1} & -(a_{1,1}+a_{1,2}+\dots+a_{1,n-1}) \\
a_{1,1}+a_{1,2} & a_{1,2}+ a_{1,3} & a_{1,3}+a_{1,4} & \dots & a_{1,n-2} + a_{1,n-1} & -(a_{1,1}+a_{1,2}+\dots+a_{1,n-2}) \\
a_{1,1}+a_{1,2}+a_{1,3} & a_{1,2}+a_{1,3}+a_{1,4} & a_{1,3}+a_{1,4}+a_{1,5} & \dots & -(a_{1,1}+a_{1,2}+\dots+a_{1,n-3}) \\
\vdots & \vdots & ⋰& ⋰ \\
a_{1,1}+a_{1,2}+\dots+a_{1,n-1} & -a_{1,1} \\
0
\end{pmatrix}
$$

Now I have the following proposition:

**Proposition.** Let $n, a_{i,j}$ be as in the setup. Then there are at least $n$ distinct pairs $(i,j)$ with $i\in\{1,\dots, n\}$ and $j\in\{1,\dots,n-i+1\}$ such that

- $a_{i,j}=0$
**or**
- $j\le n-i$ and $a_{i,j}\cdot a_{i,j+1} < 0$.

More informally, the number of zeros of the $a_{i,j}$ plus the number of "sign switches" between adjacent $a_{i,j}$ in all rows is at least $n$.

*My question: How can we prove this propostion?*.

**Context.** Proving this proposition would enable me to solve another problem about zeroes of special continuous functions that I found on StackExchange.

**Example ($n=4$).** Consider
\begin{pmatrix}
1 & \frac12 & -\frac14 & -\frac54 \\
\frac32 & \frac14 & -\frac32 \\
\frac54 & -1 \\
0
\end{pmatrix}

Then $a_{1,2}\cdot a_{1,2}<0$; $a_{2,2}\cdot a_{2,3}<0$; $a_{3,1}\cdot a_{3,2}<0$ and $a_{4,1}=0$. So in our example we have exactly $n$ zeros/sign switches.

**My work.** I tried using induction over $n$ which didn't work.

### Edit: A rigorous proof of the Proposition can be found on StackExchange. It is based on the very nice idea by Ilya Bogdanov from below.