∵a=2003x+2001，b=2003x+2002，c=2003x+2003
∴a-b=-1，b-c=-1，a-c=-2
∴a²+b²+c²-ab-ac-bc
=

1

2

(2a2+2b2+2c2−2ab−2ac−2bc )
=

1

2

[(a2−2ab+b2)+(b2−2bc+c2)+(a2−2ac+c2)]
=

1

2

[(a−b)2+(b−c)2+(a−c)2]
=

1

2

(1+1+4)
=3
故答案为3