6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from sixteen consecutive natural numbers to some number $a$ is 276, and the sum of the distances from these same sixteen numbers to some number $b$ is 748. Find all possible values of $a$, given that $a+b=62.5$.
Answer: $a=-0.75, a=46.25$.

Solution. Let the given consecutive natural numbers be denoted as $k, k+1, \ldots, k+15$. Notice that if some number lies on the segment $[k ; k+15]$, then the sum of the distances from it to the given sixteen numbers does not exceed $15 \cdot 8=120$ (the sum of the distances to the two extreme numbers is exactly 15, the sum of the distances to $k+1$ and $k+14$ does not exceed 15, the sum of the distances to $k+2$ and $k+13$ also does not exceed 15, and so on). Therefore, the numbers $a$ and $b$ lie outside the segment $[k ; k+15]$. Then the sum of the distances from the number $a$ to each of the given consecutive numbers is expressed by the formula $|16 a-k-(k+1)-\ldots-(k+15)|=16|a-k-7.5|$. Similarly, the sum of the distances from the number $b$ to each of the given numbers is $16|b-k-7.5|$. We obtain the system of equations

$$
\left\{\begin{array} { l } 
{ 1 6 | a - k - 7 , 5 | = 2 7 6 , } \\
{ 1 6 | b - k - 7 , 5 | = 7 4 8 , } \\
{ a + b = 6 2 , 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
|a-k-7.5|=17.25 \\
|b-k-7.5|=46.75 \\
a+b=62.5
\end{array}\right.\right.
$$

Consider four cases for the removal of the modulus.

1) Both numbers $a$ and $b$ lie to the right of the segment $[k ; k+15]$. Then

$$
\left\{\begin{array} { l } 
{ a - k - 7 , 5 = 1 7 , 2 5 , } \\
{ b - k - 7 , 5 = 4 6 , 7 5 , } \\
{ a + b = 6 2 , 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=16.5 \\
b=46 \\
k=-8.25
\end{array}\right.\right.
$$

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 $[k ; k+15]$. Then

$$
\left\{\begin{array} { l } 
{ - a + k + 7 , 5 = 1 7 , 2 5 , } \\
{ - b + k + 7 , 5 = 4 6 , 7 5 , } \\
{ a + b = 6 2 , 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=46 \\
b=16.5 \\
k=55.75
\end{array}\right.\right.
$$

This case also does not fit.

3) The number $a$ lies to the right, and $b$ - to the left of the segment $[k ; k+15]$. Then

$$
\left\{\begin{array} { l } 
{ a - k - 7 , 5 = 1 7 , 2 5 , } \\
{ - b + k + 7 , 5 = 4 6 , 7 5 , } \\
{ a + b = 6 2 , 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=63.25 \\
b=-0.75 \\
k=38.5
\end{array}\right.\right.
$$

This case also does not fit.

4) The number $b$ lies to the right, and $a$ - to the left of the segment $[k ; k+15]$. Then

$$
\left\{\begin{array} { l } 
{ - a + k + 7 , 5 = 1 7 , 2 5 , } \\
{ b - k - 7 , 5 = 4 6 , 7 5 , } \\
{ a + b = 6 2 , 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=-0.75 \\
b=63.25 \\
k=9
\end{array}\right.\right.
$$

Thus, it is possible only that $a=-0.75$.