$p+q+r+s+t+u=2005$ of the 6 integers, the number of integers that can be odd is 5,3 or 1. In the expression $(-1{)}^p+(-1{)}^q+(-1{)}^r+(-1{)}^s+(-1{)}^t+(-1{)}^u$

1)When one integer is even and the others are odd, then only one among is +1 and others are -1 each. Hence the sum of the terms is -5 + 1 = -4.

2)When three of the integers are even, then three of the terms are +1 each and the remaining three are -1 each. Hence the sum is 0.

3)When five of the integers are even, then five terms will be +1 each and the other term is -1. Hence the sum is 4. $\therefore$ The minimum value of the sum is -4.