I believe the answer is ab+1. We induct on p+q. Let h(p,q) be the number of minimal integers in 1,2,...,q-1 Let g(p,q) be the number of maximal integers in 1,2,...,q-1 Let s(p,q)=h(p,q)g(p,q). We will prove that s(p,q)<=q-1 for all p,q. If p>q/2 then s(p,q)=s(q-p,q) then induct. If p<q/2 let q=kp+r where k>=2. It is easy to see that g(p,kp+r)=g(p,p+r)+k-1 and h(p,kp+r)=h(p,p+r) It is easy to see that h(p,kp+r)<=p for all p,k,r Finally s(p,kp+r)=s(p,p+r)+(k-1)h(p,p+r)<=p+r-1+(k-1)(p+r)=kp+r-1, done. Please tell me if there is something wrong.

Throughout the solution, let $m,n$ replace $p,q$ and $\gcd(m,n)=1$. Let $R(x,y)$ denote the unique integer in $[0,y)$ st $y\mid x-R(x,y)$ The answer is $ab+1$. Construction: $m=b, n=ab+1$. Let $a(m,n)$ denote the number of "peaks" (we'll use peaks instead of large numbers) in $R(mt,n)$ as $t$ goes from $1$ to $n-1$. For convenience, $a(m,n)=a(m-n,n)$. Let $b(m,n)$ denote the number of valleys (we'll use valleys instead of large numbers) in $R(mt,n)$, as $t$ goes over $1$ to $n-1$. Observe $R(mt,n)+R(-mt,n)=n$ so if $t$ is a peak in the sequence $R(mt,n)$, it must be a valley in the sequence $R(-mt,n)$, and vice versa. Claim: Suppose $0<m<\frac n2$. We have $a(-m,n)=a(-m,n-m)$ and $a(m,n)=a(m,n-m)+1$. Proof: We first prove $a(m,n)=a(m,n-m)+1$. To prove this, we note $t=1,\cdots,\lfloor \frac nm \rfloor$ are all peaks. Let $S_t=\{x| x\equiv t(\bmod\; m)\}$. If I consider the sequence of $R(mt,n)$ as $t$ goes from $1$ to $n-1$, I can see I traverse $S_0, S_{-n}, S_{-2n}, \cdots$ in this order and go through each $S_i$ in ascending order. There is at most one peak in each of $S_{-n}, S_{-2n}, \cdots$ and moreover these peaks don't depend on $n$ as long as $n (\bmod\; m)$ is fixed. The proof for $a(-m,n)=a(-m,n-m)$ is analogous; notice we go through each $S_j$ in descending order, and each $|S_j|\ge 2$ so I can remove one element from each. Now we proceed by strong induction on $a+b$ to prove that the answer for $a,b$ is $ab+1$. The base case, $(1,1)$ is trivial. Let $f(a,b)$ denote our answer. Note $f(b,a)=f(a,b)$ because if $a(m,n)=a, b(m,n)=b$ then it must follow $a(-m,n)=b, b(-m,n)=a$. Suppose $m$ satisfies $a(m,f(a,b))=a, b(m,f(a,b))=b$. Suppose $m<\frac{f(a,b)}{2}$ or we replace $m$ with $-m$ and focus on $f(b,a)=f(a,b)$. In this case, $m\ge b$ because the number of valleys is at most $m$ because only $1,\cdots,m$ are eligible to be valleys. Therefore, $f(a,b)-m$ satisfies $a(m,f(a,b)-m)=a-1, b(m,f(a,b)-m)=b$, so $f(a,b)\ge m+f(a-1,b)\ge b+(a-1)b+1=ab+1$, completing the induction.