6. For any natural $n$ and for any set of numbers $x_{1}, x_{2}, \ldots, x_{n}$ from the interval $[0 ; 3]$, the equation $\sum_{i=1}^{n}\left|x-x_{i}\right|=a n$ has a solution $x$ belonging to the interval $[0 ; 3]$. Indicate which of the following values of $a$ satisfy this condition:
a) $a=0$, b) $a=\frac{3}{2}$, c) $a=2$.
Answer: only $a=\frac{3}{2}$. Solution. To prove the impossibility of any $a$ except $a=\frac{3}{2}$, we can provide an example: $n=2, x_{1}=0, x_{2}=3$. Then $\sum_{i=1}^{n}\left|x-x_{i}\right|=|x-0|+|x-3|=3$ for any $x \in[0 ; 3]$, and the resulting equation $3=2 a$ has no solutions for all $a$ except for $a=\frac{3}{2}$. Now we will prove that solutions always exist when $a=\frac{3}{2}$.

Consider the function $f(x)=\frac{1}{n} \cdot \sum_{i=1}^{n}\left|x-x_{i}\right|$. Then $f(0)=\frac{1}{n} \cdot \sum_{i=1}^{n}\left|-x_{i}\right|=\frac{1}{n} \cdot \sum_{i=1}^{n} x_{i}$, $0 \leq f(0) \leq 3$.

$$
f(3)=\frac{1}{n} \cdot \sum_{i=1}^{n}\left|3-x_{i}\right|=\frac{1}{n} \cdot \sum_{i=1}^{n}\left(3-x_{i}\right)=\frac{1}{n} \cdot\left(3 n-\sum_{i=1}^{n} x_{i}\right)=3-f(0), 0 \leq f(3) \leq 3
$$

Thus, $f(0)+f(3)=3$, with one of the terms being no less than $\frac{3}{2}$, and the other no more than $\frac{3}{2}$, and $f(x)$ is continuous on the interval $[0 ; 3]$. Therefore, there must be a value of $x$ for which $f(x)=\frac{3}{2}$.

Answer for variant 152: only $a=2$.

Answer for variant 153: only $a=\frac{5}{2}$.

Answer for variant 154: only $a=3$.