9.4 At a round table, 20 shareholders are seated. What is the minimum possible value of the total number of their shares, given that a) any three of them together have more than 1000 shares, b) any three consecutive shareholders together have more than 1000 shares?
Answer: a) 6679; b) 6674. Solution. a) Let $x_{i}$ be the number of shares owned by the $i$-th shareholder (if counted in a circle, starting from some shareholder). Arrange the numbers $x_{i}$ in ascending order, we get: $y_{1} \leq y_{2} \leq y_{3} \leq \ldots \leq$ $y_{20}$. Note that $y_{3} \geq 334$, because otherwise the sum $y_{1}+y_{2}+y_{3}$ would be $\leq 999$. Therefore, for the sum $S$ of all shares, we have: $S \geq\left(y_{1}+y_{2}+y_{3}\right)+334 \cdot 17 \geq 1001+334 \cdot 17=6679$. This value is achieved when one shareholder has 333 shares, and all the others have 334 shares. b) Writing inequalities of the form $x_{i}+x_{i+1}+x_{i+2} \geq 1001$ for all $i$ (assuming that $x_{21}=x_{1}, x_{22}=x_{2}$) and adding all 20 inequalities, we get $3 S \geq$ $1001 \cdot 20$. Thus, $S \geq 6673.333 \ldots$ i.e. $S \geq 6674$. This value is achieved when the numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{19}$, $x_{20}$ are $333,334,334, \ldots$ (the triplet repeats 6 times) and the last two numbers are $334,334$.