The answer is yes just take : $\left ( a,b,c,d \right )=(100^{33},2.100^{33},3.100^{33},4.100^{33})$

I solved this both for $\mathbb N,\mathbb Z$ For $\mathbb N,$let $10^{66}|a,b,c,d$ for simplicity.Then we need $a^3+b^3+c^3+d^3=10^2,$using the identity ${1^3+2^3+....+n^3=\frac {n^2(n+1)^2} 4}^2,$we get $(a_1,b_1,c_1,d_1)=(1,2,3,4)$ and all its permutation, since then $a_1+b_1+c_1+d_1=10$.Then the numbers are $10^{66},2.10^{66},3.10^{66},4.10^{66}$ Now for $\mathbb Z,$we take $a=5.10^{66},b=c=10^{66},d=-3.10^{66}$