Is it possible to put $+$ or $-$ signs in such a way that $\pm 1 \pm 2 \pm \cdots \pm 100 = 101$?

I'm reading a book about combinatorics. Even though the book is about combinatorics there is a problem in the book that I can think of no solutions to it except by using number theory.

Problem: Is it possible to put $+$ or $-$ signs in such a way that $\pm 1 \pm 2 \pm \cdots \pm 100 = 101$?

My proof is kinda simple. Let's work in mod $2$. We'll have:

$\pm 1 \pm 2 \pm \cdots \pm 100 \equiv 101 \mod 2$ but since $+1 \equiv -1 \mod 2$ and there are exactly $50$ odd numbers and $50$ even numbers from $1$ to $100$ we can write:

$(1 + 0 + \cdots + 1 + 0 \equiv 50\times 1 \equiv 0) \not\equiv (101\equiv 1) \mod 2$ which is contradictory.

Therefore, it's not possible to choose $+$ or $-$ signs in any way to make them equal.

Now is there a combinatorial proof of that fact except what I have in mind?

Solution 1:

You can rephrase essentially the same argument in the following terms:

Suppose that there were such a pattern of plus and minus signs. Let $P$ be the set of positive terms, and let $N$ be the set of negative terms together with the number $101$. Then $\sum P-\sum N=0$, so $\sum P=\sum N$, and $\{P,N\}$ is a partition of $\{1,2,\ldots,101\}$ into two sets with equal sum. But $\sum_{k=1}^{101}k=\frac12\cdot101\cdot102=101\cdot51$ is odd, so this is impossible.

Solution 2:

Another answer that use almost the same idea: the sum or subtraction of two even or odd number is an even number. How many odd number we have?