, 2022

The least possible number for $q$ is $ab + 1$.
Let $S_{p,q}$ be the collection of small numbers, while $T_{p,q}$ be the collection of big numbers. Write out $S_{p,q} \cup T_{p,q}$ in increasing order, and we get a sequence $c_1, \dots, c_{a+b-1}$ (note that clearly $S_{p,q} \cap T_{p,q} = \{1\}$). We claim that for each $n \ge 2$, if $i, j < n$ are the maximum indices such that $c_i \in S_{p,q}$, $c_j \in T_{p,q}$, then $c_n = c_i + c_j$, and furthermore $c_n \in S_{p,q}$ if and only if $c_i p \bmod q > (-c_j p) \bmod q$, in which case $c_n p \bmod q = (c_i p \bmod q) - (-c_j p \bmod q)$.
We will show this by induction on $n$. When $n=2$ this is trivial. Now suppose that it holds for all indices smaller than $n$. By replacing $p$ with $-p$, we can WLOG assume that $c_i p \bmod q > (-c_j p) \bmod q$.
If $c_n < c_i + c_j$ and $c_n \in T_{p,q}$, then $(-c_n p) \bmod q < (-c_j p) \bmod q$, showing that $0 < (c_n - c_j) p \bmod q < (-c_j p) \bmod q < c_i p \bmod q$. Since $c_n - c_j < c_i$, this contradicts the fact that $c_i \in S_{p,q}$.
If $c_n < c_i + c_j$ and $c_n \in S_{p,q}$, then we have $c_n - c_i < c_j$ and $0 < -c_n c_j p \bmod q < c_i p \bmod q$. Let $c_k$ be the largest index smaller than $c_j$ that lies in $T_{p,q}$. Then it is clear that $(-c_k p) \bmod q < c_i p \bmod q$. If $i < j$, then by the inductive hypothesis, we have $c_{i+1} \in S_{p,q}$, which contradicts the choice of $i$. Therefore $i > j$. Now let $\ell < j$ be the largest such that $c_\ell \in S_{p,q}$. Then since $\ell < j < i$, by the inductive hypothesis we have $(-c_j p) \bmod q = (-c_k p \bmod q) - (c_\ell p \bmod q) < (c_i p \bmod q) - (c_i p \bmod q) = 0$. This is a contradiction.
Therefore $c_n \ge c_i + c_j$. However, note that $(c_i + c_j) \bmod q = (c_i p \bmod q) - ((-c_j p) \bmod q) < c_i p \bmod q$, showing that $c_i + c_j \in S_{p,q}$ and so $c_n = c_i + c_j$.
Hence we can construct the sequence $c_n$ as follows: we first start with a red positive integer $x$ and a blue positive integer $y$. Each time we choose the last red number and the last blue number and add them together, and color the number red or blue. We write $m$ red numbers and $n$ blue numbers in total. The situation above is when $x = y = 1, m = a - 1, n = b - 1$ and red numbers are members of $S_{p,q}$, blue numbers are members of $T_{p,q}$. Let $f(m, n, x, y)$ be the minimum possible total of the last written red number and the last written blue number. Since in the original setting, the last red number $s$ and the last blue number $t$ clearly satisfies $sp \bmod q = (-tp) \bmod q = 1$ and so $s + t = q$, what we need to do is to give a lower bound of $f(a, b, 1, 1)$. In fact, we can see that if $m, n \ge 1$, then
$$f(m, n, x, y) = \min(f(m - 1, n, x + y, y), f(m, n - 1, x, x + y))$$
where the first term corresponds to coloring the next number red, and the second term corresponds to coloring the next number blue. We also have that $f(m, 0, x, y) = x + (m + 1)y$, $f(0, n, x, y) = (n + 1)x + y$. Thus, we can induct on $m + n$ and show that $f(m, n, x, y) = mn \min(x, y) + (m + 1)x + (n + 1)y$: suppose that it holds for all smaller $m + n$, then
$$\begin{align*} f(m, n, x, y) \\ &= \min \left( (m-1)ny + m(x+y) + (n+1)y, m(n-1)x + (m+1)x + (n+1)(x+y) \right) \\ &= mn \min(x, y) + (m+1)x + (n+1)y, \end{align*}$$
as desired. Thus, $q \ge f(a-1, b-1, 1, 1) = ab + 1$.

To show that it is possible that $q = ab + 1$, pick $p = a$. Then $T_{p,q} = \{1, 2, \dots, b\}$ and $S_{p,q} = \{1, b+1, \dots, (a-1)b+1\}$.