Consider the matrix $A=\begin{pmatrix}1-x_1^{-1}&-x_2^{-1}&\ldots&-x_n^{-1}\\ -x_1^{-1}&1-x_2^{-1}&\ldots&-x_n^{-1}\\ \vdots&\vdots&\ddots&\vdots\\ -x_1^{-1}&-x_2^{-1}&\ldots&1-x_n^{-1}\end{pmatrix}$ obtained by the one above dividing the $j$-th column by $x_j$. Then, let $u=(1,1,\ldots,1)^t$, so that $Au=\begin{pmatrix}1-x_1^{-1}-x_2^{-1}-\ldots-x_n^{-1}\\ -x_1^{-1}+1-x_2^{-1}-\ldots-x_n^{-1}\\ \vdots\\ -x_1^{-1}-x_2^{-1}-\ldots+1-x_n^{-1}\end{pmatrix}=\begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}\;.$ Hence, $u\in\ker A$, so $\mathrm{rk}\; A\leq n-1$.
On the other hand, set $D=\mathrm{diag}(x_1,\ldots, x_n)$ and $U=(u_{ij})$ with $u_{ij}=1$ for every $1\leq i,j\leq n$. Then $A+U=D$ and $\mathrm{rk}U=1$, $\mathrm{rk}D=n$, hence $\mathrm{rk} A\geq n-1$. [Edit: this is essentially the comment by hardmath and it's proved by noticing that $\mathbb{R}^n=\mathrm{Im}A+U\subseteq\mathrm{Im}A \oplus \mathrm{Im}U$ and, as $\mathrm{Im}U$ is one-dimensional, $\mathrm{Im}A$ has to be at least $n-1$-dimensional.]