Polish A.

Chichikov is playing with Nozdryov. First, Nozdryov distributes 1001 nuts among three boxes. After looking at the distribution, Chichikov names any integer \( N \) from 1 to 1001. Then Nozdryov must, if necessary, move one or several nuts to an empty fourth box and show Chichikov one or several boxes where the total number of nuts is exactly \( N \). As a result, Chichikov will receive as many dead souls as the number of nuts Nozdryov moved. What is the maximum number of souls Chichikov can guarantee for himself, regardless of how Nozdryov plays?
Upper bound. By placing 143, $286=2 \cdot 143$ and $572=4 \cdot 143$ nuts in the boxes, Nosdriv can, for any $N$, move no more than 71. Indeed, $N$ can be represented as $143 k+r$, where $0 \leq k \leq 7$, and $-71 \leq r < 71$. The number $143 k$ can be collected with one or several boxes without moving anything. If $r < 0$, then $1001-N=143(7-k)-r$. By moving $r$ nuts, we will get several boxes with $1001-N$ nuts. Then in the remaining boxes, there will be $N$ nuts, which we will present.

Lower bound. We will show that for any arrangement, there is an $N$ that will require moving no fewer than 71 nuts. Let $x, y$ and $z$ nuts lie in the boxes. The six numbers $x, y, z, x+y, x+z, y+z$ divide the large segment $[0,1001]$ into seven smaller ones (some of which may be degenerate). Among them, there is a segment of length not less than $1001: 7$ = 

143. On this segment, there is an integer that is at least 71 away from the ends of the segment. Without moving, we can only obtain sets where the total number of nuts lies at the end of one of the seven small segments. To change this number by $r$, it is necessary to move no fewer than $r$ nuts.

## Answer

71 souls.