$Q(x)=x^3-24x^2+143x$. $P(x)=x(x-120)(x-240)$. The nine roots are $0,1,3,5,8,11,13,15,16$.

Let $z_1,z_2,z_3$ be the three roots of the polynomial $P(x)=(x-z_1)(x-z_2)(x-z_3)$, and let $Q(x)=x^3-sx^2+tx-u$. Then the nine roots of \[ P(Q(x)) ~=~ (Q(x)-z_1) \, (Q(x)-z_2)\, (Q(x)-z_3) \] are the roots $a_i,b_i,c_i$ of $Q(x)-z_i$ where $i$ runs from $1$ to $3$. The Viet\'e relations yield that for $1\le i\le3$ we have \[ a_i+b_i+c_i = s\\ a_ib_i+b_ic_i+a_ic_i = t \] Since the sum of all nine roots is $72$, this implies $s=24$. Hence we are looking for three disjoint triples of non-negative integers that add up to $24$ and for which the sum of squares is the same (since $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)$). A little bit of experimentation shows that choosing $t=143$ works with $a_1=0$, $b_1=11$, $c_1=13$ and $a_2=1$, $b_2= 8$, $c_2=15$ and $a_3=3$, $b_3= 5$, $c_3=16$. Hence the two polynomials $Q(x) = x^3-24x^2+143x$ and $P(x) = x(x-120)(x-240)$ have all the desired properties, and the answer to the problem is YES.