Let's assume that by removing a finite number of positive elements from the set of natural numbers, we obtained a set $S$ that is closed under addition. Let $k$ be an element of $S$. How many elements of $S$ are there such that subtracting $k$ from them results in a number that does not belong to $S$?
Solution. Consider the residue classes modulo $k$ of the natural numbers. (A number $x$ is in the $i$-th class $(0 \leq i \leq k-1)$ if it gives a remainder of $i$ when divided by $k$.) Since $S$ is obtained from the set of natural numbers by omitting a finite number of elements, we have omitted only a finite number of elements from each residue class. Therefore, $S$ contains elements from each of the residue classes modulo $k$.

Let us select the smallest element from each residue class, denoted by $x_{i}$ for the $i$-th residue class. Since $S$ is closed under addition, $x_{i}+k, x_{i}+2k$, etc., are elements of $S$. Subtracting $k$ from any of these elements thus yields an element of the $i$-th residue class and hence an element of $S$. Only $x_{i}-k \notin S$.

Since $S$ is divided into $k$ residue classes, it has $k$ elements from which subtracting $k$ results in a number not belonging to $S$.