6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from seven consecutive natural numbers to some number $a$ is 609, and the sum of the distances from these same seven numbers to some number $b$ is 721. Find all possible values of $a$, given that $a+b=192$.
Answer: $a=104, a=191, a=1$.

Solution. Let the given consecutive natural numbers be denoted by $k, k+1, \ldots, k+6$. Notice that if some number lies on the interval $[k ; k+6]$, then the sum of the distances from it to the given seven numbers does not exceed $\frac{7}{2} \cdot 6=21$ (the sum of the distances to the two extreme numbers is exactly 6, the sum of the distances to $k+1$ and $k+5$ does not exceed 6, the sum of the distances to $k+2$ and $k+4$ also does not exceed 6, and the distance to $k+3$ does not exceed half the length of the interval between the extreme numbers, i.e., 3). Therefore, the numbers $a$ and $b$ lie outside the interval $[k ; k+6]$. Then the sum of the distances from the number $a$ to each of the given consecutive numbers is expressed by the formula $|7 a-k-(k+1)-\ldots-(k+6)|=7|a-k-3|$. Similarly, the sum of the distances from the number $b$ to each of the given numbers is $7|b-k-3|$. We obtain the system of equations

$$
\left\{\begin{array} { l } 
{ 7 | a - k - 3 | = 6 0 9 , } \\
{ 7 | b - k - 3 | = 7 2 1 , } \\
{ a + b = 1 9 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
|a-k-3|=87 \\
|b-k-3|=103 \\
a+b=192
\end{array}\right.\right.
$$

Consider four cases for the removal of the modulus.

1) Both numbers $a$ and $b$ lie to the right of the interval $[k ; k+6]$. Then

$$
\left\{\begin{array} { l } 
{ a - k - 3 = 8 7 , } \\
{ b - k - 3 = 1 0 3 , } \\
{ a + b = 1 9 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=88 \\
b=104 \\
k=-2
\end{array}\right.\right.
$$

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

2) Both numbers $a$ and $b$ lie to the left of the interval $[k ; k+6]$. Then

$$
\left\{\begin{array} { l } 
{ - a + k + 3 = 8 7 , } \\
{ - b + k + 3 = 1 0 3 , } \\
{ a + b = 1 9 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=104 \\
b=88 \\
k=188
\end{array}\right.\right.
$$

3) The number $a$ lies to the right, and $b$ to the left of the interval $[k ; k+6]$. Then

$$
\left\{\begin{array} { l } 
{ a - k - 3 = 8 7 , } \\
{ - b + k + 3 = 1 0 3 , } \\
{ a + b = 1 9 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=191 \\
b=1 \\
k=101
\end{array}\right.\right.
$$

4) The number $b$ lies to the right, and $a$ to the left of the interval $[k ; k+6]$. Then

$$
\left\{\begin{array} { l } 
{ - a + k + 3 = 8 7 , } \\
{ b - k - 3 = 1 0 3 , } \\
{ a + b = 1 9 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=1 \\
b=191 \\
k=85
\end{array}\right.\right.
$$

Thus, there are three possible cases: $a=1, a=191, a=104$.