1. The sum of 86 consecutive integers is 2021. By how much is the arithmetic mean of all positive numbers among them greater than the arithmetic mean of all negative numbers?
1. The sum of 86 consecutive integers is 2021. By how much is the arithmetic mean of all positive numbers among them greater than the arithmetic mean of all negative numbers?

## Solution.

Let the smallest of these 86 numbers be $x$. The others are $x+1, x+2, \ldots, x+85$.

We have:

$x+(x+1)+(x+2)+\ldots+(x+85)=2021$

$x+x+1+x+2+\ldots+x+85=2021$

$86 x+1+2+\ldots+85=2021$.

The sum $1+2+\ldots+85$ is

$1+2+\ldots+85=86 \cdot 85: 2=3655$.

Further, we have:

$86 x+3655=2021$

$86 x=-1634$

$x=-19$

The desired numbers are $-19,-18, \ldots, 66$.

The sum of all positive numbers among them is

$1+2+\ldots+66=67 \cdot 66: 2=2211$,

and their arithmetic mean is $2211: 66=33.5$.

The sum of all negative numbers among them is

$-(1+2+\ldots+19)=-20 \cdot 19: 2=-190$,

and their arithmetic mean is $-190: 19=-10$.

The desired difference is $33.5-(-10)=43.5$.