Let's assume that $S = 0$. Let's choose an arbitrary division of the string into two parts. It is clear that the sum of the numbers in the two parts is $0$, so one of them is not less than $0$, and the other is not greater than $0$. If they are not $0$, we get a contradiction, so the sum of the numbers of any smaller string is $0$, so all the numbers in the string are $0$, which is what we need to prove.

Let $S \neq 0$, without restriction of generality let $S > 0$. Let $a$ be any number in the string. Consider an arbitrary division of the string into two parts: either both sums are not less than $0$, or not greater than $0$. It is clear that the second option is impossible, because the sum of two nonnegative integers cannot be equal to $S > 0$. Then the sum of the numbers of any lesser row is not less than $0$.

Let $x$ and $y$ denote the sum of the numbers to the left and right of $a$ respectively (if there are no such numbers, then we assume the corresponding variable is $0$). Then, from the above, $x, y \geq 0$ and $x + a, y + a \geq 0$, therefore $x + y \geq 0$ and $x + y + 2a \geq 0$. By definition, $S = x + y + a$, so $S - a \geq 0$ and $S + a \geq 0$, i.e. $S \leq a \leq S$, therefore we get $|a| \leq S$.