In fact, equation has no more than $[a_1,a_2,...,a_k]$ solutions in positive integers. Solution: $n=b_i (mod a_i)$ for all $1\leq i\leq k$ $[\frac{n}{a_1}]+[\frac{n}{a_2}]+...+[\frac{n}{a_k}]=n$ $=\frac{n-b_1}{a_1}+\frac{n-b_2}{a_2}+...+\frac{n-b_k}{a_k}$ $=n.(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_k})-(\frac{b_1}{a_1}+\frac{b_2}{a_2}+...+\frac{b_k}{a_k})$ $n.(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_k}-1)=\frac{b_1}{a_1}+\frac{b_2}{a_2}+...+\frac{b_k}{a_k}$ If equation has more than $[a_1,a_2,...,a_k]$ integer solution, there exist different $n_1$ and $n_2$ such that given same $b_i$, so $n_1=n_2$ Contradiction.