I think the answer is no. It's easy to prove that if a and b are coprime positive integers then any integer which is at least ab can be written as ax + by,where x and y are nonnegative integers.Also if a and b would not be coprime then any multiple of (a;b) which is at least [a;b] can be written as ax + by. Assume such a set exists.Then any two elements must not be coprime.Also if x is an element and d is a divisor of x,then there must be a finite number of y such that (x;y)=d.But x has a finite number of divisors,so we get a contradiction.

Exist,For example E=(1,4,7)

IMO2022Goldinshallah wrote: Exist,For example E=(1,4,7) This set is not a correct example. Because $7=4+1+1+1$. parmenides51 wrote: Does there exist a subset $E$ of the set $N$ of all positive integers such that none of the elements in $E$ can be presented as a sum of at least two other (not necessarily distinct) elements from $E$ ? (E. Barabanov) However the answer is indeed yes. We can take as an example $E=\{5,14,22\}$. Obviously $5,14$ cannot be represented as a sum of at least two elements of $E$. if $22$ can be represented as a sum of at least two elements of $E$, then it has to be of the form $5x+14y=22$. Where $x,y$ is a natural number. But this implies $y=1$, which implies $5x=8$ but this is not possible. So $E$ indeed satisfies the condition. Edit: @below Yes, it is possible.

EmilXM wrote: However the answer is indeed yes.. This is correct, however I think that the original (and the more interesting one) question might have been to have such a set but which is infinite, to which #2 correctly responded negatively.