44. There are $2 k+1$ cards, numbered with consecutive natural numbers from 1 to $2 k+1$. What is the maximum number of cards that can be selected so that no one of the selected numbers is equal to the sum of two other selected numbers?
44. The cards that satisfy the condition of the problem have numbers $k+1, k+2, \ldots, 2 k+1$. Their number is $k+1$. We will prove that it is impossible to choose a greater number of cards. Suppose the opposite, i.e., that it is possible to choose $k+r$ cards that satisfy the condition of the problem, where $r>1$. Let $n$ be the largest number written on these cards. Consider the differences between $n$ and the other numbers on the selected cards. There are $k+r-1$ such differences, and they must be written on the remaining $2 k+1-(k+r)$ cards, i.e., the inequality

$$
k+r-1 \leqslant 2 k+1-k-r=k-r+1
$$

or $r \leqslant 1$ must hold. Contradiction.