As this question has not been answered for a very long time, I think it is appropriate to use a strong conjecture to resolve it, namely Schinzel's hypothesis H. Let us write $P(x)=P_1(x)^{k_1}...P_r(x)^{k_r}$ where $P_i(x)\in \mathbb{Z}[x]$ are irreducible, distinct, non-constant, and have positive leading coefficient (possibly $r=1$). Let the constant coefficient of $P_i$ be $c_i$. We may assume that $c_i$ are all non-zero, since otherwise we arrive at the trivial case $x\mid P(x)$. We have
\begin{align*}
\phi(n)\mid \phi(P(n))= \phi(P_1(n)^{k_1}...P_r(n)^{k_r})
\end{align*}
for all $n\geq 1.$ Let $c=|c_1...c_r|$, and define the polynomials $Q_i(x)=\frac{P_i(cx)}{|c_i|}\in \mathbb{Z}[x]$. Then we have
\begin{align*}
\phi(cn)\mid \phi(Q_1(n)^{k_1}...Q_r(n)^{k_r}|c_1|^{k_1}...|c_r|^{k_r}).
\end{align*}
for all $n\geq 1$. The advantage of this was that $Q_i$ have their constant coefficients equal to $\pm 1$. Let $D$ be large enough, in particular $D>\deg P$ (but we impose another condition later). We have in particular
\begin{align*}
\phi(cD!n)\mid \phi(Q_1(D!n)^{k_1}...Q_r(D!n)^{k_r}|c_1|^{k_1}...|c_r|^{k_r}).
\end{align*}
for all $n$, and these new polynomials have no small prime divisors. Finally, we apply Schinzel's hypotheisis H to the irreducible polynomials $x,Q_1(D!x),...,Q_r(D!x)$ (by Gauss' lemma, $f(x)$ is irreducible in $\mathbb{Z}[x]$ if and only if $f(kx)$ is). Their product does not have any fixed prime divisor $q$ because then we would have
\begin{align*}
\prod_{i=1}^r Q_i(D!x)\equiv 0 \mod q
\end{align*}
for $x=1,2,...,q-1$. However, this conguence has at most $\deg P$ solutions, so $q\leq \deg P+1$, which is a contradiction by the definition of $D$ (none of the polynomials $Q_i(D!x)$ is identically zero $\pmod q$ since their constant coefficients are $\pm 1$). Hence, we have infinitely many primes $p$ such that $Q_1(D!p),...,Q_r(D!p)$ are all primes. Setting $n=p$ in the previous formula involving $n$, we obtain by multiplicativity for large enough $p$ that
\begin{align*}
p-1\mid Q_1(D!p)^{k_1-1}(Q_1(D!p)-1)...(Q_r(D!p))^{k_r-1}(Q_r(D!p)-1)\phi(|c_1|^{k_1}...|c_r|^{k_r}).
\end{align*}
Since $Q_i(D!p)\equiv Q_i(D!)\pmod{p-1}$, we get
\begin{align*}
p-1\mid Q_i(D!)^{k_1-1}(Q_1(D!)-1)...Q_r(D!)^{k_r-1}(Q_r(D!)-1)\phi(|c_1|^{k_1}...|c_r|^{k_r}).
\end{align*}
However, we can choose $D$, depending only on the polynomial $P$, so that $Q_i(D!)\geq 2$ for all $i$, and then the right-hand side of the previous divisibility relation should be smaller than the left-hand side, which is a contadiction for $p$ large enough.
I doubt that this problem can be solved without assuming some conjecture as $\phi(n)$ is surprisingly hard to control for general $n$; for example, Lehmer's totient problem about solving $\phi(n)\mid n-1$ seems perhaps easier but is known to be open.
So, to gain some control, one would like to choose $n$ in the relation $\phi(n)\mid \phi(P(n))$ to be prime, or at least closely related to them. However, very little is known about prime values of polynomials, and even less about their simultaneous prime values. In fact, we do not even know (according to this paper by M.Filazeta) a polynomial $f(x)\in \mathbb{Z}[x]$ such that $\deg f\geq 4$ and $f(x)$ is square-free for infinitely many integers $x$, although most numbers are squarefree and any polynomial with no fixed prime divisor should work. There are results about polynomials having infinitely almost prime values, but I am not sure whether they would be very helpful here.
I think I got the proof that no such real polynomial with degree $ \geq 6$ exists.
Let $n \geq 6$
Suppose for contradiction that $z_1,\ldots,z_n \in \mathbb R-\{0\}^n$ are such that $(X-z_1)...(X-z_n)=X^n+\sum_{k=1}^{n-1}z_iX^{n-i}$
Then three useful identities appear $$\sum_{k=1}^{n}z_k=-z_1 \; \; \; \;(1)$$
$$\sum_{\large1\leq i<j \leq n}z_iz_j=z_2 \; \; \; \;(2)$$
$$\prod_{k=1}^n z_k=(-1)^n z_n \; \; \; \;(3)$$
Since $$(\sum_{k=1}^{n}z_k)^2=\sum_{k=1}^{n}z_k^2+2\sum_{\large1\leq i<j \leq n}z_iz_j$$it follows that$$z_1^2=2z_2+\sum_{k=1}^{n}z_k^2$$
Hence $$0< \sum_{k=2}^{n}z_k^2=-2z_2 \; \; \; \;(4)$$ and $$0<\sum_{k=3}^{n}z_k^2=1-(z_2+1)^2 \; \; \; \;(5) $$
$(4)$ and $(5)$ imply $$\; \; \; \;-2<z_2<0 \; \; \; \;(6)$$
thus $(6)$ and $(4)$ imply $$0<\sum_{k=2}^{n}z_k^2 < 4 \; \; \; \; (7)$$
Also $(6)$ and $(5)$ imply $$0<\sum_{k=3}^{n}z_k^2 \leq 1 \; \; \; \; (8)$$
By AM-GM, $$\left(|z_3|^2\ldots|z_{n-1}|^2 \right)^{1/(n-3)} \leq \frac{1}{n-3}\sum_{k=3}^{n-1}z_k^2 \leq \frac{1}{n-3}\sum_{k=3}^{n}z_k^2$$
Hence
$$|z_3|^2\ldots|z_{n-1}|^2 \leq \left(\frac{1}{n-3}\sum_{k=3}^{n}z_k^2\right)^{n-3} $$
Squaring, $$|z_3|\ldots|z_{n-1}| \leq \left(\frac{1}{n-3}\sum_{k=3}^{n}z_k^2\right)^{\large \frac{n-3}{2}} \leq_{ \large (8)} \dfrac{1}{{(n-3)}^{(n-3)/2}} \; \; \; \; (9)$$
By triangle inequality $(1)$, and Cauchy-Schwarz
$$2|z_1| \leq \sum_{k=2}^{n}|z_k| \leq \sqrt{n-1} \sqrt{\sum_{k=2}^{n}z_k^2} $$
Hence by $(7)$,
$$|z_1| \leq \sqrt{n-1} \; \; \; \; (10)$$
Rewriting $(6)$ as $$|z_2|\lt2 \; \; \; \; (11) $$
Recalling $(3)$ (with $z_n$ cancelled from both sides) and putting together $(9)$, $(10)$ and $(11)$, we have
$$1=|z_1||z_2||z_3|\cdots|z_{n-1}| < \dfrac{ 2\sqrt{n-1}}{{(n-3)}^{(n-3)/2}}$$
This inequality fails for $n\geq 6$.
Contradiction.
I can't prove anything for $n=5$ so maybe the conjecture doesn't hold.
Best Answer
Hint: the number of $(a,b,c,d)\in\left(\mathbb{Z}^+\right)^4$ such that $a+b+c+d=9$ is given by the coefficient of $x^9$ in the product $(x+x^2+x^3+\ldots)^4$, i.e. by $$ [x^9]\frac{x^4}{(1-x)^4} = [x^5]\frac{1}{(1-x)^4}=[x^5]\sum_{n\geq 0}\binom{n+3}{3}x^n =\binom{5+3}{3}=\color{red}{56}.$$ See also stars and bars.