Zhendarov R.G.

On a plane, $n$ lines are drawn. Each intersects exactly with 1999 others. Find all $n$ for which this is possible.
Let's divide all the drawn lines into groups of parallel lines. Let there be $k$ such groups. It is clear that in each group there are $n-1999$ lines. Therefore,

$n=k(n-1999)$, that is, $1999 k=n(k-1)$. Since the numbers $k$ and $k-1$ are coprime, 1999 (which is a prime number) divides $k-1$. Thus, $k-1=1999$ or $k-1=1$. Accordingly, $n=2000$ or 3998.

## Answer

2000 or 3998.