Solution: ∫ Integer x 1, x2, x3, …, x2008 satisfies: ①- 1≤xn≤2, n= 1, 2, …, 2008; ②x 1+x2+…+x 2008 = 2008； ③x 12+x22+…+x20082=2008，

∴x 1=x2=x3=…=x2008= 1，

∴x 13+x23+…+x20083=2008，

The minimum value of x13+x23+…+x20083 is 2008, and the maximum value is 2008.

So the answer is: 2008,2008.