Proizvoov V.v.

Ten consecutive natural numbers were written on the board. When one of them was erased, the sum of the nine remaining numbers turned out to be 2002.

What numbers remained on the board?
Let $x$ be the smallest of the written numbers. Denote by $x+y$ the crossed-out number ($0<y<9$). Then $x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8)+(x+9)-(x+y)=2002$, that is, $9x=1957+y. 1957+y$ is divisible by 9 only when $y=5$. Therefore, $x=1962: 9=218$.

## Answer

$218,219,220,221,222,224,225,226$ and 227.