Simple addition and subtraction

For every 3 terms, there would be the following combination:

1-2 ... +3

4-5 ... +6

7-8 ... +9 and so on...

2001÷3=667

The whole expression will become:

-1(667)+(3×1+3×2+3×3+3×4+...+3×667)+2002-2003

=-677+3[ $\frac{667(668)}{2}$ ]+(-1)

= $\boxed{667666}$

Notice that:

$1-2+3=2\\ 4-5+6=5\\ 7-8+9=8\\ \ldots\\ 1999-2000+2001=2000$

From here, we can know that the sum:

$1-2+3+4-5+6+\ldots+1999-2000+2001$

is actually the sum of the arithmetic progression:

$2+5+8+\ldots+2000$

with $a=2, d=3$ and $T_n=2000$

Use the formula to find the value of $n$ :

$T_n = a + (n-1)(d)\\ 2000 = 2 + (n-1)(3)\\ \implies n=667$

Make use of the formula $S_n = \dfrac{n}{2}(a+T_n)$ to find the value of the expression:

$1-2+3+4-5+6+\ldots+1999-2000+2001+2002-2003\\ =(2+5+8+\ldots+2000) +(2002-2003)\\ =\dfrac{667}{2}(2+2000) -1\\ =(667)(1001)-1\\ =667667-1\\ =\boxed{667666}$