6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from nine consecutive natural numbers to some number $a$ is 294, and the sum of the distances from these same nine numbers to some number $b$ is 1932. Find all possible values of $a$, given that $a+b=256$.
Answer: $a=\frac{13}{3}, a=\frac{755}{3}$.

Solution. Let the given consecutive natural numbers be denoted by $k, k+1, \ldots, k+8$. Notice that if some number lies on the segment $[k ; k+8]$, then the sum of the distances from it to the given nine numbers does not exceed $\frac{9}{2} \cdot 8=36$ (the sum of the distances to the two extreme numbers is exactly 8, the sum of the distances to $k+1$ and $k+7$ does not exceed 8, the sum of the distances to $k+2$ and $k+6$ also does not exceed 8, and so on; the distance to $k+4$ does not exceed half the length of the segment between the extreme numbers, i.e., 4). Therefore, the numbers $a$ and $b$ lie outside the segment $[k ; k+8]$. Then the sum of the distances from the number $a$ to each of the given consecutive numbers is expressed by the formula $|9 a-k-(k+1)-\ldots-(k+8)|=9|a-k-4|$. Similarly, the sum of the distances from the number $b$ to each of the given numbers is $9|b-k-4|$. We obtain the system of equations

$$
\left\{\begin{array} { l } 
{ 9 | a - k - 4 | = 2 9 4 , } \\
{ 9 | b - k - 4 | = 1 9 3 2 , } \\
{ a + b = 2 5 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
|a-k-4|=32 \frac{2}{3} \\
|b-k-4|=214 \frac{2}{3} \\
a+b=256
\end{array}\right.\right.
$$

Consider four cases for the removal of the modulus.

1) Both numbers $a$ and $b$ lie to the right of the segment $[k ; k+8]$. Then

$$
\left\{\begin{array} { l } 
{ a - k - 4 = 3 2 \frac { 2 } { 3 } } \\
{ b - k - 4 = 2 1 4 \frac { 2 } { 3 } , } \\
{ a + b = 2 5 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=37 \\
b=219 \\
k=\frac{1}{3}
\end{array}\right.\right.
$$

Since $k$ must be a natural number, this case does not fit.

2) Both numbers $a$ and $b$ lie to the left of the segment $[k ; k+8]$. Then

$$
\left\{\begin{array} { l } 
{ - a + k + 4 = 3 2 \frac { 2 } { 3 } , } \\
{ - b + k + 4 = 2 1 4 \frac { 2 } { 3 } , } \\
{ a + b = 2 5 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=219 \\
b=37 \\
k=247 \frac{2}{3}
\end{array}\right.\right.
$$

This case also does not fit.

3) The number $a$ lies to the right, and $b$ lies to the left of the segment $[k ; k+8]$. Then

$$
\left\{\begin{array} { l } 
{ a - k - 4 = 3 2 \frac { 2 } { 3 } , } \\
{ - b + k + 4 = 2 1 4 \frac { 2 } { 3 } , } \\
{ a + b = 2 5 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=251 \frac{2}{3} \\
b=4 \frac{1}{3} \\
k=215
\end{array}\right.\right.
$$

4) The number $b$ lies to the right, and $a$ lies to the left of the segment $[k ; k+8]$. Then

$$
\left\{\begin{array} { l } 
{ - a + k + 4 = 3 2 \frac { 2 } { 3 } , } \\
{ b - k - 4 = 2 1 4 \frac { 2 } { 3 } , } \\
{ a + b = 2 5 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=4 \frac{1}{3}, \\
b=251 \frac{2}{3}, \\
k=33
\end{array}\right.\right.
$$

Thus, there are two possible cases: $a=4 \frac{1}{3}=\frac{13}{3}, a=251 \frac{2}{3}=\frac{755}{3}$.