\section*{Problem 1 - 311011}

Determine the number of all pairs \((x ; y)\) of positive natural numbers \(x\) and \(y\) for which the following inequality (1) holds:

\[
x+y<1991
\]
}

For \(x=1\), (1) is satisfied exactly for \(y=1,2, \ldots, 1988,1989\); hence, there are

1989 pairs with \(x=1\) for which (1) holds.

Similarly, one obtains a total of

1988 pairs with \(x=2\) for which (1) holds,

1987 pairs with \(x=3\) for which (1) holds, ...

1 pair with \(x=1989\) for which (1) holds, namely the pair \((1989,1)\)

The total number of pairs for which (1) holds is thus equal to the sum 1+...+1988+1989; this is, according to a well-known formula,

\[
\frac{1989 \cdot 1990}{2}=1979055
\]