Given two positive integers $p$ and $q$, we are to determine the smallest number $n$ such that the operation of choosing two identical numbers $a, a$ on the blackboard and replacing them with $a+p$ and $a+q$ can go on infinitely.

To solve this, we first note that we can assume $\gcd(p, q) = 1$ by scaling, because the problem is invariant under scaling by the greatest common divisor.

We claim that the smallest $n$ is $\frac{p+q}{\gcd(p, q)}$. When $\gcd(p, q) = 1$, this simplifies to $p + q$.

To see that $n = p + q$ is sufficient, consider a board with the set $\{1, \dots, p\} \cup \{1, \dots, q\}$. This configuration can last forever under the given operation.

We now show that $n \ge p + q$ is necessary. Assume $n$ is minimal, which implies that every entry is changed infinitely many times. We consider the entire blackboard as generating an infinite table with $n$ columns, such that each row is obtained from the previous one by replacing $a, a$ with $a+p, a+q$ (for some $a$), and each column is unbounded.

Without loss of generality, we can assume (by shifting and rearranging) that the first two entries of the first row are $0$, and all others are nonnegative. We add the condition that whenever the first column is erased, we increment that entry by $p$, and whenever the second column is erased, we increment that entry by $q$. Thus, the first column will contain all positive multiples of $p$ and the second column will contain all positive multiples of $q$.

Claim: Let $S = \{ p, 2p, \dots, (q-1)p \} \cup \{ q, 2q, \dots, (p-1)q \}$. Then for every $s \in S$, there exists a column $C$ other than the first or second column such that $\max (S \cap C) = s$.

Proof: Let $t \in S$ and assume $p \mid t$ (the other case is similar). Since it is incremented by $p$ in the first column, there must be some column containing $t$ followed immediately by $t+q$. That column then cannot contain any larger elements of $S$. Indeed, the next smallest multiples of $p$ and $q$ exceeding $t+q$ are $t+pq$ and $pq+q$, respectively. $\blacksquare$

Hence, the number of columns is at least $2 + \# S = p + q$, as needed.

The answer is $\boxed{\frac{p+q}{\gcd(p,q)}}$.