jmc

By Cauchy-Schwarz, $$(a_1+a_2+\cdots +a_n)(a_1^3+a_2^3+\cdots +a_n^3)\ge (a_1^2+a_2^2+\cdots +a_n^2{)}^2.$$Since $96\cdot 216=144^2,$ we have equality in the Cauchy-Schwarz Inequality, which means $$\frac{a_1^3}{a_1}=\frac{a_2^3}{a_2}=\cdots =\frac{a_n^3}{a_n}.$$Then $a_1^2=a_2^2=\cdots =a_n^2,$ so $a_1=a_2=\cdots =a_n.$

From the given, $n{a}_1=96$ and $n{a}_1^2=144.$ Dividing these equations, we get $a_1=\frac{3}{2},$ so $n=\boxed{64}.$