Counting number of points in a lattice with bounded sup norm Let $\Lambda$ be a lattice in $\mathbb{R}^n$. For $\bar{x} \in \mathbb{R}^n$, let 
$\| \bar{x} \| = max_{1 \leq i \leq n} \{ |x_i| \}$, i.e. the sup norm. Let $\lambda_1, ..., \lambda_n$
be a successive minima of $\Lambda$ with respect to the sup norm. 
I am interested in counting number of points 
$N(U) = \# \{ \bar{x} \in \Lambda : \| \bar{x} \| < U  \}$ when say $\lambda_{j} \leq U < \lambda_{j+1}$. I am expecting some like
$$
\frac{U^j}{\lambda_1 ... \lambda_j}  \ll N(U) \ll \frac{U^{j}}{\lambda_1 ... \lambda_{j}}
$$ 
when $\lambda_{j} \leq U < \lambda_{j+1}$, where the 
implicit constants may depend on $\det (\Lambda)$ and $n$, but not on $\Lambda$ itself.
I would appreciate any assistance, comments, or references. Thank you very much!
PS And I mean the inequalities as they are. Thank you!
 A: Your expectation is correct. Let $\lambda_{j} \leq U < \lambda_{j+1}$. There are independent lattice vectors $v_1,\dots,v_j\in\Lambda$ such that $\|v_i\|=\lambda_i$ for $1\leq i\leq j$. 
Consider the linear combinations
$ c_1v_1+\dots +c_j v_j\in\Lambda$ with integral coefficients $c_i\in\mathbb{Z}$ satisfying $|c_i|\leq U/(2n\lambda_i)$. Their sup-norms are $\leq U/2<U$, and their number is $\gg_n U^j/(\lambda_1\dots\lambda_j)$. This proves the lower bound. 
Now consider the subspace $V_j=\mathbb{R}v_1+\dots+\mathbb{R}v_j$ equipped with the $j$-volume coming from the euclidean norm in $\mathbb{R}^n$, i.e. an orthonormal basis in $V_j$ spans a $j$-cube of $j$-volume $1$. Then, $\Lambda_j=\Lambda\cap V_j$ is a lattice in $V_j$ with successive minima $\lambda_1,\dots,\lambda_j$, hence by Minkowski's second theorem its fundamental parallelepiped $P_j$ has $j$-volume
$$\text{vol}_j(P_j)\geq 2^{-j}\cdot\lambda_1\dots\lambda_j\cdot\text{vol}_j(B_j),$$
where $B_j=\{ \bar{x} \in V_j : \| \bar{x} \| \leq 1  \}.$
The sup-norm is majorized by the euclidean norm, hence $B_j$ contains a euclidean $j$-ball of radius $1$, which shows that $\text{vol}_j(P_j)\gg_j \lambda_1\dots\lambda_j$.
On the other hand,
$$\{ \bar{x} \in \Lambda : \| \bar{x} \| < U  \}=\{ \bar{x} \in \Lambda_j : \| \bar{x} \| < U  \},$$ 
whence the translates of $P_j$ by the elements of this set are pairwise disjoint and they all lie in $\{ \bar{x} \in V_j : \| \bar{x} \| < (j+1)U  \}$ whose $j$-volume is $\ll_j U^j$. This proves the upper bound.
Remark. The above argument was enhanced and corrected by the OP's comments, for which I am grateful.
