**14**

n^3+8=(n+2)*(n^2-2n+4) -----> (1)

n^2-4=(n+2)*(n-2) -----> (2)

(1)/(2)= (n^2-2n+4)/(n-2)= n+ 4/(n-2) -----> (3)

For (3) to be an integer 4/(n-2) must be integer, then n-2 = {-4,-2,-1,1,2,4} => n={-2,0,1,3,4,6}

n=-2 is excluded as n^2-4 can't be zero due to divisibility rule

Sum of all n will be 0+1+3+4+6=**14**