@orthodontist has already given a clean and educated answer to my question. Nevertheless let me include an elementary answer, tweaking Vijayaraghavan's construction. The calculations are not terribly ugly after all.

Fix integers
$$
\ell_1>\cdots>\ell_{n-k}>0>\ell_{n-k+1}>\cdots>\ell_n
$$
whose sum is $0$ and such that $\ell_i-\ell_{i+1}\ge 2$ for each $i\in\{1,\dots,n-1\}$. Let $e_0:=0$ and $e_i:=\ell_1+\cdots+\ell_i$ for $i\in\{1,\dots,n\}$. Let $b \ge 3$ be another integer. I claim that the polynomial
$$
P(x) := \sum_{i=0}^n (-1)^i b^{e_i} x^{n-i}
$$
has the required properties $(\star)$ (except perhaps for irreducibility).
To begin, note that $e_i \ge 0$ for each $i$ and so $P$ has integer coefficients. Furthermore, it is monic, has constant term $(-1)^n$.

In order to locate the roots of $P$, fix integers $c_1, \dots, c_{n-1}$ such that $$
\ell_1>c_1>\ell_2>c_2>\cdots>c_{n-1}>\ell_n \quad \text{and} \quad c_{n-k}=0.
$$
Fix $j\in\{1,\dots,n-1\}$. I claim that in the expression
$$
P(b^{c_j}) = \sum_{i=0}^n (-1)^i b^{e_i + (n-i)c_j} ,
$$
the dominant term corresponds to $i=j$. Indeed, the sequence of integers $$i \in \{0,\dots,n\} \mapsto e_i + (n-i)c_j$$ is increasing for $i$ in the interval
$\{0,\dots,j\}$, and decreasing afterwards. Using the fact that $2\sum_{r=1}^\infty b^{-r} \le 1$ (i.e. $b \ge 3$) we conclude that **the sign of $P(b^{c_j})$ is $(-1)^j$**. By the intermediate value theorem, $P$ contains a root in each of the intervals
$$
(0, b^{c_{n-1}}), \ (b^{c_{n-1}},b^{c_{n-2}}), \ \dots, \ (b^{c_2},b^{c_1}), \ (b^{c_1},+\infty) \, .
$$
Since $b^{c_{n-k}}=1$, we conclude that $P$ has $k$ simple roots on the interval $(0,1)$ and $n-k$ simple roots on the interval $(1,+\infty)$.

EDIT: **Getting irreducibility** (Thanks @GabeConant for pointing the error of the previous "argument".)

Let $z_1>\dots>z_n$ be the roots of $P$. The construction actually ensures that:
$$
b^{\ell_i - 1} < z_i < b^{\ell_i + 1} \quad \text{for each } i.
$$
Obviously, $\prod_{i=1}^n z_i = 1$.

**Assume that $P$ is reducible.**
Then there exists a nonempty proper subset $I \subset \{1,\dots,n\}$ such that:
$$
\prod_{i \in I} z_i = 1.
$$
(Indeed, if $P=QR$ is a non-trivial factorization of $P$ then $\prod_{Q(z)=0} z$ and $\prod_{R(z)=0} z$ must be positive integers and so must equal $1$.)
Therefore:
$$
\left| \sum_{i \in I} \ell_i \right| < n \, .
$$

Now, we have lots of freedom in the choice of the numbers $\ell_i$, and it is certainly possible (though no clean argument occurs to me right now) to choose them so that $\left| \sum_{i \in I} \ell_i \right| \ge n$ for each nonempty proper subset $I \subset \{1,\dots,n\}$. In this way we can guarantee that $P$ is irreducible.