1. Let's say that three mutually distinct natural numbers form a sum triplet if the sum of the first two of them equals the third number. Determine the maximum number of sum triplets that can be found in a set of twenty natural numbers.
1. For any twenty natural numbers

$$
x_{1}<x_{2}<\ldots<x_{20}
$$

we estimate how many sum triples, i.e., triples $\left\{x_{i}, x_{j}, x_{k}\right\}$ satisfying the conditions $1 \leqq i<j<k \leqq 20$ and $x_{i}+x_{j}=x_{k}$, can exist, first for a fixed index $k \in \{3,4, \ldots, 20\}$. Let these triples be $\left\{x_{i_{1}}, x_{j_{1}}, x_{k}\right\},\left\{x_{i_{2}}, x_{j_{2}}, x_{k}\right\}, \ldots,\left\{x_{i_{p}}, x_{j_{p}}, x_{k}\right\}$. Then the numbers

$$
x_{i_{1}}, x_{j_{1}}, x_{i_{2}}, x_{j_{2}}, \ldots, x_{i_{p}}, x_{j_{p}}
$$

are all distinct and lie in the set $\left\{x_{1}, x_{2}, \ldots, x_{k-1}\right\}$, so for their count $2 p$ we have the estimate $2 p \leqq k-1$, from which $p \leqq\left[\frac{1}{2}(k-1)\right]$ (where $[a]$ denotes the integer part of the number $a$). Therefore, the total number of all sum triples cannot be greater than the sum

$$
\sum_{k=3}^{20}\left[\frac{k-1}{2}\right]=1+1+2+2+3+3+\ldots+9+9=90
$$

The example of the set $M=\{1,2, \ldots, 20\}$ shows that the number 90 of sum triples is achievable, because for each $k \in\{3,4, \ldots, 20\}$ we can choose any number from the set $\left\{1,2, \ldots,\left[\frac{1}{2}(k-1)\right]\right\}$ for the number $i$; the corresponding integer $j=k-i$ then indeed satisfies the inequalities $i<j<k$, so $\{i, j, k\}$ is a sum triple lying in $M$.

For a complete solution, award 6 points. If the solver provides an example of a suitable set of 20 numbers and explains why it contains 90 sum triples, but does not justify that a larger number is not possible (it is not enough to write that it is obvious!), award 3 points.