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
Notice that every polynomial satisfying the two conditions must be positive on $\mathbb{R}$, therefore all its roots are negative. Now Let's write $P=X\prod_{1\leq k \leq n-1}(X+a_{i})$
we have $(\sum_{1 \leq k \leq n-1} a_{k})((\sum_{1 \leq k \leq n-1} \frac1{a_{k}})\geq (n-1)^2$, therefore $(\sum_{1 \leq k \leq n-1} a_{k})((\sum_{1 \leq k \leq n-1} (\prod_{1 \leq j \leq n-1} a_{j})\frac1{a_{k}})\geq (n-1)^2(\prod_{1 \leq j \leq n-1} a_{j})$
Notice that the coefficient of $x^{n-1} $(respectively $x$) in the expansion of $P=\prod_{1\leq k \leq n-1}(x+a_{i})$ is $(\sum_{1 \leq k \leq n-1} a_{k})$ (respectively $(\sum_{1 \leq k \leq n-1} (\prod_{1 \leq j \leq n-1} a_{j})\frac1{a_{k}})$ ). therefore there product is inferior to $n(n-1)$
but this implies $ \frac{n}{n-1}\geq (\prod_{1 \leq j \leq n-1} a_{j}) $ therefore being the coefficient of $x^{1}$ in $P$ the quantity $ (\prod_{1 \leq j \leq n-1} a_{j}) =1 or 2$. it equals 2 if and only if n=2 i.e $P=X(X+2)$ .
I think that if you push the reasoning and consider the coefficient of $x^{k}$ times the coefficient of $x^{n-k}$ you can prove that this is the only solution for $n\geq2$.