Clearly, because of the cardinality stuff, $n$ should be divisible by $3.$ Now we will show that for $n\ge 6$ and divisible there exist a construction [ Note in these constructions, $A, B, C$ denotes the three disjoint set ] For $n=6,$ consider this construction $$1^A~~2^B~~3^C$$$$4^C~~5^B~~6^A$$ And note that this construction can be used for any $6$ consecutive integers i.e for $k,k+1,k+2,k+3,k+4,k+5$ consider this:- $$k^A~~~{k+1}^B~~{k+2}^C$$$${k+3}^C~~{k+4}^B~~{k+5}^A$$ And for $n=9,$ consider this $$1^A~~2^B~~3^C$$$$4^C~~5^A~~6^B$$$$7^B~~8^C~~9^A$$ Now we are set for making the construction for any $n=3k, k>1.$ Now if $3k$ is even, then we can just repeat our construction for $n=6.$ [ like grouping them in consecutive $6$ integers i.e $(1-6),(7-12),\dots $] And if $3k$ is odd, then consider $n=3k-9.$ Since $3k-9$ is even, we can do the construction, we did for even and then the last $9$ integers can be arranged in the $n=9$ construction. And we are done!