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?
 A: 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?
A: If $T_n=n(n+1)/2$ is the $n^{th}$ triangular number, an inductive proof (using $T_n+(n+1)=T_{n+1}$) shows the attainable numbers at step $n$ are
$$-T_n,\ -T_n+2,\ \cdots , T_n-2, \ T_n,$$ 
in particular they all have the same parity as $T_n$. Since $T_{100}=5050$ is even, we see that $101$ cannot be attained in any way by $100$ steps.
addendum: The first triangular number at least $101$ is $T_{14}=105$ (and is odd). This overshoots the goal $101$ by $4$, so if we take the sum
$$1+2+3+\cdots+14=105$$
and change the sign on the $2$, we get $101$. Seems this is the only way to get $101$ in 14 steps, and we cannot get it with 13 or fewer since $T_{13}=91$ is the largest with $13$ steps.
A: Replacing 100 with $n$
and using Brian M. Scott's solution,
we want a partition of
$\{1, 2, ..., n+1\}$
into two sets with equal sums.
The sum is
$\frac{(n+1)(n+2)}{2}$,
and if $n=4k$,
this is
$(4k+1)(2k+1)$
which is odd
and therefore impossible.
If $n = 4k+1$,
this is
$(2k+1)(4k+3)$
which is also odd,
and therefore impossible.
If $n = 4k+2$,
this is
$(4k+3)(2k+2)$,
so it is not ruled out,
and each sum must be
$(4k+3)(k+1)$.
if $n = 4k+3$,
this is
$(2k+2)(4k+5)$
which is also not ruled out,
and each sum must be
$(k+1)(4k+5)$.
Now I'll try to find a solution
for the not impossible cases.
(I am working these out as I enter them.)
For the $n=4k+2$ case,
the sum must be
$(4k+3)(k+1)
=(4k+4-1)(k+1)
=4(k+1)^2-(k+1)
=(2k+2)^2-(k+1)
$.
The square there suggests,
to me,
the formula for
the sum of consecutive odd numbers
$1+3+...+(2m-1)=m^2$,
so $1+3+...+(4k+3) = (2k+2)^2$.
If $k+1$ is odd,
remove it from the sum
so it is
$(2k+2)^2-(k+1)$.
If $k+1$ is even,
both $1$ and $k$ are odd,
so remove them from the sum.
In either case, we have the desired partition.
For the $n=4k+3$ case,
the sum must be
$(4k+5)(k+1)
=(4k+4+1)(k+1)
=4(k+1)^2+(k+1)
=(2k+2)^2+(k+1)
$.
Again,
$1+3+...+(4k+3) = (2k+2)^2$.
If $k+1$ is even,
add it to the sum
so it is
$(2k+2)^2+(k+1)$.
If $k+1$ is odd,
$k+2$ is even,
so remove $1$
and add $k+2$ to the sum.
In either case, we have the desired partition.
I do not know if these partitions
are unique.
A: 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.
A: Consider both sides modulo $2$. Then the right side is $1$, whereas the left side is $0$ (since it consists of $50$ ones and $50$ zeros).
