jmc

Recall that $$(1+2+3+\dots +n{)}^2=1^3+2^3+3^3+\dots +n^3.$$ Thus we have that for $n\ge 7$, $(1+2+3+\dots +n{)}^2=1^3+2^3+3^3+\dots +n^3\ge 1^3+2^3+\dots +7^3$, while $(1+2+3+\dots +6{)}^2=1^3+2^3+3^3+\dots +6^3$, which is less than the desired sum. So our answer is $\boxed{6}.$