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The answer is $no$.

In the plane, the circle of diameter $d$ greater than $1$ but smaller than $\sqrt2$ and centered at $({1\over2},{1\over2})$ contains no point with integer coordinates, and the square $K_2$ circumscribed about it with diagonals parallel to the coordinate axes contains no such point either. Let $K_3\subset\mathbb{R}^3$ be the Cartesian product $K_2\times[0,d]$. Then $K_3$ contains no lattice point, the $2$-sphere inscribed in $K_3$ is of diameter $d$, and the number of facets of $K_3$ is $6$, since, in general, the product $K\times[0,1]$ has two more facets than $K$. Iterating the product procedure, we get $K_n=K_2\times[0,d]^{n-2}\subset\mathbb{R}^n$ with $2n$ facets, containing no lattice point but containing a sphere of radius $d>1$.

Similarly, a sequence of polytopes $K_{k,n}$ ($k$ fixed, $n=k, k+1, k+2, \ldots$) can be constructed that have the same properties as above, with $d$ arbitrarily close to $\sqrt k$. This shows that for any constant $d$ in place of $1$ in the question asked here, the answer is still in the negative. Namely, begin with $K_k$ to be the cross-polytope in $\mathbb{R}^k$ circumscribed about the $k$-sphere of diameter slightly smaller than $\sqrt k$ and centered at $({1\over2},{1\over2}, \ldots,{1\over2})$ and with main diagonals (the "cross") parallel to the coordinate axes, then proceed as above. The number of facets of $K_{k,n}$ will be $2^k + 2(n-k)$, which is $O(2n)$.

The answer is $no$.

In the plane, the circle of diameter $d$ greater than $1$ but smaller than $\sqrt2$ and centered at $({1\over2},{1\over2})$ contains no point with integer coordinates, and the square $K_2$ circumscribed about it with diagonals parallel to the coordinate axes contains no such point either. Let $K_3\subset\mathbb{R}^3$ be the Cartesian product $K_2\times[0,d]$. Then $K_3$ contains no lattice point, the $2$-sphere inscribed in $K_3$ is of diameter $d$, and the number of facets of $K_3$ is $6$, since, in general, the product $K\times[0,1]$ has two more facets than $K$. Iterating the product procedure, we get $K_n=K_2\times[0,d]^{n-2}\subset\mathbb{R}^n$ with $2n$ facets, containing no lattice point but containing a sphere of diameter $d>1$.

Similarly, a sequence of polytopes $K_{k,n}$ ($k$ fixed, $n=k, k+1, k+2, \ldots$) can be constructed that have the same properties as above, with $d$ arbitrarily close to $\sqrt k$. This shows that for any constant $d$ in place of $1$ in the question asked here, the answer is still in the negative. Namely, begin with $K_k$ to be the cross-polytope in $\mathbb{R}^k$ circumscribed about the $k$-sphere of diameter slightly smaller than $\sqrt k$ and centered at $({1\over2},{1\over2}, \ldots,{1\over2})$ and with main diagonals (the "cross") parallel to the coordinate axes, then proceed as above. The number of facets of $K_{k,n}$ will be $2^k + 2(n-k)$, which is $O(2n)$.

The answer is "no".

In the plane, the circle of diameter $d$ greater than $1$ but smaller than $\sqrt2$ and centered at $({1\over2},{1\over2})$ contains no point with integer coordinates, and the square $K_2$ circumscribed about it with diagonals parallel to the coordinate axes contains no such point either. Let $K_3\subset\mathbb{R}^3$ be the Cartesian product $K_2\times[0,d]$. Then $K_3$ contains no lattice point, the $2$-sphere inscribed in $K_3$ is of diameter $d$, and the number of facets of $K_3$ is $6$, since, in general, the product $K\times[0,1]$ has two more facets than $K$. Continuing inductively, we get $K_n=K_2\times[0,d]^{n-2}\subset\mathbb{R}^n$ with $2n$ facets, containing no lattice point but containing a sphere of radius $d>1$.

Similarly, a sequence of polytopes $K_{k,n}$ ($k$ fixed, $n=k, k+1, k+2, \ldots$) can be constructed that have the same properties as above, with $d$ arbitrarily close to $\sqrt k$. This shows that for any constant $d$ in place of $1$ in the question asked here, the answer is still in the negative. Namely, begin with $K_k$ to be the cross-polytope in $\mathbb{R}^k$ circumscribed about the $k$-sphere of diameter slightly smaller than $\sqrt k$ and centered at $({1\over2},{1\over2}, \ldots,{1\over2})$ and with main diagonals (the "cross") parallel to the coordinate axes, then proceed as above. The number of facets of $K_{k,n}$ will be $2^k + 2(n-k)$, which is $O(2n)$.

The answer is $no$.

In the plane, the circle of diameter $d$ greater than $1$ but smaller than $\sqrt2$ and centered at $({1\over2},{1\over2})$ contains no point with integer coordinates, and the square $K_2$ circumscribed about it with diagonals parallel to the coordinate axes contains no such point either. Let $K_3\subset\mathbb{R}^3$ be the Cartesian product $K_2\times[0,d]$. Then $K_3$ contains no lattice point, the $2$-sphere inscribed in $K_3$ is of diameter $d$, and the number of facets of $K_3$ is $6$, since, in general, the product $K\times[0,1]$ has two more facets than $K$. Iterating the product procedure, we get $K_n=K_2\times[0,d]^{n-2}\subset\mathbb{R}^n$ with $2n$ facets, containing no lattice point but containing a sphere of radius $d>1$.

Similarly, a sequence of polytopes $K_{k,n}$ ($k$ fixed, $n=k, k+1, k+2, \ldots$) can be constructed that have the same properties as above, with $d$ arbitrarily close to $\sqrt k$. This shows that for any constant $d$ in place of $1$ in the question asked here, the answer is still in the negative. Namely, begin with $K_k$ to be the cross-polytope in $\mathbb{R}^k$ circumscribed about the $k$-sphere of diameter slightly smaller than $\sqrt k$ and centered at $({1\over2},{1\over2}, \ldots,{1\over2})$ and with main diagonals (the "cross") parallel to the coordinate axes, then proceed as above. The number of facets of $K_{k,n}$ will be $2^k + 2(n-k)$, which is $O(2n)$.

The answer is "no". In the plane, the circle of diameter greater than $1$ but smaller than $\sqrt2$ and centered at $({1\over2}, {1\over2})$ contains no point with integer coordinates, and the square circumscribed about it with a horizontal diagonal contains no such point either. Similar examples exist in every dimension, with a suitable $n$-sphere instead of a circle and the cross-polytope with diameters parallel to the coordinate axes, circumscribed about the sphere in place of the square. The diameter of the $n$-sphere can be very close to $\sqrt n$, and the number of vertices is $2n$.

The answer is "no". 

In the plane, the circle of diameter $d$ greater than $1$ but smaller than $\sqrt2$ and centered at $({1\over2},{1\over2})$ contains no point with integer coordinates, and the square $K_2$ circumscribed about it with diagonals parallel to the coordinate axes contains no such point either. Let $K_3\subset\mathbb{R}^3$ be the Cartesian product $K_2\times[0,d]$. Then $K_3$ contains no lattice point, the $2$-sphere inscribed in $K_3$ is of diameter $d$, and the number of facets of $K_3$ is $6$, since, in general, the product $K\times[0,1]$ has two more facets than $K$. Continuing inductively, we get $K_n=K_2\times[0,d]^{n-2}\subset\mathbb{R}^n$ with $2n$ facets, containing no lattice point but containing a sphere of radius $d>1$.

Similarly, a sequence of polytopes $K_{k,n}$ ($k$ fixed, $n=k, k+1, k+2, \ldots$) can be constructed that have the same properties as above, with $d$ arbitrarily close to $\sqrt k$. This shows that for any constant $d$ in place of $1$ in the question asked here, the answer is still in the negative. Namely, begin with $K_k$ to be the cross-polytope in $\mathbb{R}^k$ circumscribed about the $k$-sphere of diameter slightly smaller than $\sqrt k$ and centered at $({1\over2},{1\over2}, \ldots,{1\over2})$ and with main diagonals (the "cross") parallel to the coordinate axes, then proceed as above. The number of facets of $K_{k,n}$ will be $2^k + 2(n-k)$, which is $O(2n)$.