Sorry for my bad English.
Let $X$ be projective variety over algebraically closed field $k$,
and $L_1,\dots, L_n$ be any line bundles on $X$.
Now can we construct very ample line bundle $M$ on $X$ such that $M\otimes L_i$ is also very ample for $1\le i\le n$?
(In Mumford's "Abelian varieties" p.145 of new edition, he says this is possible.)
Best Answer
Let $X$ be your projective variety, and let $\mathcal O(1)$ be a very ample sheaf on $X$ associated to a closed immersion of $X$ into $\mathbb P_k^r$.
Because $\mathcal O(1)$ is ample (since it is very ample) and invertible sheaves are coherent, for every $i$ you have that there exists an integer $m_i > 0$ such that for all $m > m_i$, the sheaf $\mathcal L_i \otimes \mathcal O(m)$ is generated by global sections (here, $\mathcal L_i$ is the invertible sheaf corresponding to the line bundle $L_i$)
Let now $M$ be some integer larger than all the $m_i$. Then for all $i$, the sheaves $\mathcal L_i \otimes \mathcal O(M)$ are generated by global sections.
By exercise II.7.5d) in Hartshorne's Algebraic Geometry (see for example this post), the sheaves $\mathcal L_i \otimes \mathcal O(M + 1)$ are very ample.
Thus, if you take $B$ to be the very ample line bundle associated to the very ample sheaf $\mathcal O(M + 1)$, you have that $B \otimes L_i$ is very ample for all $i$.