China Girls' Mathematical Olympiad

For any two vectors $\mathbf{x}=(x_1,x_2,\dots ,x_n)$ and $\mathbf{y}=(y_1,y_2,\dots ,y_n)$, if there exists $1\le k\le n$ such that $$x_1=y_1,\dots ,x_{k-1}=y_{k-1},x_k>y_k,$$ we then denote it as $\mathbf{x}>\mathbf{y}$. Let $\mathbf{x}=(x_1,x_2,\dots ,x_n)$ represent the distribution of the balls among the boxes. Then $\mathbf{x}$ is a non-negative integer vector. The operation defined in the question, if executable, can be expressed as $\mathbf{x}+{\mathbf{\alpha }}_k$, where ${\mathbf{\alpha }}_1=(-1,1,0,\dots ,0)$, ${\mathbf{\alpha }}_k=(0,\dots ,0,1,-2,1,0,\dots ,0)$ ($2\le k\le n-1$), ${\mathbf{\alpha }}_n=(0,\dots ,0,1,-1)$. Then for $k\ge 2$, we always have $\mathbf{x}+{\mathbf{\alpha }}_k>\mathbf{x}$. So for any initial distribution of the balls, after a finite number of operations on every $B_k$ ($k\ge 2$) that contains at least two balls, we can arrive at a ball distribution $\mathbf{y}=(y_1,y_2,\dots ,y_n)$ satisfying $y_k\le 1$ for all $k\ge 2$. If at this time $y_2=\cdots =y_n=1$, the problem is solved; otherwise, we have $y_1\ge 2$.

Assuming $i$ is the smallest number such that $y_i=0$, we can then do a series of operations on $B_1,B_2,\dots ,B_{i-1}$: $$\begin{array}{rl}& (y_1,1,\dots ,1,0,y_{i+1},\dots ,y_n)\stackrel{B_1,B_2,\dots ,B_{i-1}}{\to }\\ & (y_1,1,\dots ,1,0,1,y_{i+1},\dots ,y_n)\stackrel{B_1,B_2,\dots ,B_{i-2}}{\to }\\ & (y_1,1,\dots ,1,0,1,1,y_{i+1},\dots ,y_n)\to \cdots \to \\ & (y_1,0,1,\dots ,1,y_{i+1},\dots ,y_n)\stackrel{B_1}{\to }\\ & (y_1-1,1,\dots ,1,1,y_{i+1},\dots ,y_n)\end{array}$$ to get $(y_1-1,1,\dots ,1,y_{i+1},\dots ,y_n)$. Repeating the operations above, we can finally arrive at the ball distribution vector that meets the requirement. $\square$