National Math Olympiad in Slovenia

First, let us show the following: if the number on the blackboard has decreased after the change, then the new number is greater than or equal to $5$. We cannot obtain the number $1$ because the sum of two positive integers can never be equal to $1$. The number $2 = 1 + 1$ can only be obtained from $1$, the number $3 = 1 + 2$ can only be obtained from $2$, and the number $4 = 1 + 3 = 2 + 2$ can only be obtained from $3$ or $4$. So, if the number obtained is less than $5$, then it has not decreased during the change. If the starting number was $n < 5$, then we cannot get a smaller number.

Now, let us show that if a number is greater than $5$, then it can be reduced. Assume that the number $m > 5$ is on the blackboard at a given time. If $m$ is even, i.e. of the form $m = 2k$, where $k > 2$ is a positive integer, then it can be replaced by $k + 2$. In this way we have obtained a number smaller than $m$, since the condition $k + 2 < 2k$ is equivalent to the assumption $k > 2$. If $m$ is odd, then we can express it as $m = 2k - 1$, where $k > 3$ is a positive integer. In this case, we can first replace it by $(2k - 1) + 1 = 2k$ and then with $k + 2$. Once again we have obtained a number smaller than $m$, because the condition $k + 2 < 2k - 1$ is equivalent to our assumption $k > 3$. Therefore, if the starting number is $n \ge 5$, we can reduce it to the number $5$, using a finite number of steps, and according to the reasoning above this is the smallest number that can be obtained.

The smallest number we can obtain after a finite number of steps is equal to $5$ if $n \ge 5$ and is equal to $n < 5$.