VMO

a) We first observe that one cannot write the number $2013$ between a pair $(a,b)$ with $a+b>2013$. Using this simple observation, it is easy to check that the number $2013$ is written only twice after $2013$ steps in the $8$th and the $1013$th steps.

b) We add an extra number $1$ after $1000$ on the line. We perform $2013$ such steps on this new line and count how many times the number $2013$ is written. We construct a sequence $A$ of pairs inductively as follows. Initially, $$A=\{(1,2),(2,3),\dots ,(999,1000),(1000,1)\}.$$ In each step of the transformation, whenever we write a number $k=a+b$ between the pair $(a,b)$ (from left to right), we add two pairs $(a,k)$ and $(k,b)$ into the sequence $A$. We will prove by induction on $a+b$ that an ordered pair $(a,b)$ appears on the sequence $A$ only if $\text{gcd}(a,b)=1$, and if $\text{gcd}(a,b)=1$ then the pair $(a,b)$ appears exactly once on the sequence $A$. Suppose that this statement holds for any $a+b<k$, we show that it also holds when $a+b=k$. Without loss of generality, we can assume that $a<b$ (the case $a>b$ is similar). The pair $(a,b)$ is added to the sequence $A$ whenever we write the number $b$ between the pair $(a,b-a)$. Since $a+(b-a)=b<k$, the pair $(a,b-a)$ appears on the sequence $A$ only if $\text{gcd}(a,b-a)=1$, and if $\text{gcd}(a,b-a)=1$ then the pair $(a,b-a)$ appears exactly once on the sequence $A$. Since $$\text{gcd}(a,b)=\text{gcd}(a,b-a),$$ the statement holds for $a+b=k$. By the induction principle, the statement holds for any $(a,b)$. This implies that the number of times that $2013$ is written is the number of pairs $(a,2013-a)$ with $\text{gcd}(a,2013-a)=1$. Therefore, the number $2013$ is written $\varphi (2013)=1200$ times on the new line.

By the part a), the number $2013$ is written twice between $1000$ and $1$. Hence, the number $2013$ is written $1200-2=1198$ times on the given line. $\square$