Two valued variant of subset sum problem
Your problem $P$: given a multiset $S = \{a_1, \dots, a_m\}$ of some $m_1$ elements equal to $a \in \mathbb{Z}^+$ and $m_2$ elements equal to $b \in \mathbb{Z}^+$, and $k \in \mathbb{N}$, what is the $k$-partition $S_1, \dots, S_k$ which maximizes $\min_i \sum_{a_j \in S_i} a_j$.
Definition A &quot;disjoint collection&quot; of subsets of $S$ is a collection of subsets of $S$ that are disjoint. A &quot;disjoint $k$-collection&quot; is a disjoint collection with $k$ subsets.
Consider problem $P'$: given a multiset $S$ of $m_1$ elements equal to $a \in \mathbb{Z}^+$ and $m_2$ elements equal to $b \in \mathbb{Z}^+$, and $T \in \mathbb{Z}^+$, what is the maximally sized disjoint collection of subsets of $S$ which has the property that each subset's sum is $\ge T$?
Claim 1 if we can solve $P'$ in poly-time we can solve $P$ in poly-time.
Proof of Claim 1: We reduce $P$ to $P'$ using binary search. Let $A'$ denote the assumed poly-time algorithm for $P'$.
So we have $S$ and some $k$ as input from $P$. Let $T^*$ denote the optimal minimum-set-sum (the value we're trying to find to answer $P$).
We will call $A'$ with the same $S$ and various $T$s as input.
Let $\tilde{T}$ be the maximal $T \in \mathbb{Z}^+$ for which $A'$ returns a collection of size at least $k$.
We claim that $\tilde{T} = T^*$.
First, let's show $\tilde{T} \le T^*$: to see this, by definition of $\tilde{T}$, note that $A'$ produces a set of at least $k$ subsets each of whom has sum $\ge \tilde{T}$; add the &quot;leftover&quot; elements outside to any subset in the collection; now we have a $k$-partition with each subset having sum $\ge \tilde{T}$. But $T^*$ is the maximal $T$ with this property: so $T^* \ge \tilde{T}$.
Next, let's show $\tilde{T} \ge T^*$: by definition of $T^*$, there is some $k$-partition of $S$ all of whose subsets sum to $\ge T^*$. So if we feed $S$, $T^*$ as input to $A'$, $A'$ will return a collection of size at least $k$. But $\tilde{T}$ is the maximal such $T$, so $\tilde{T} \ge T^*$.
We know that $\tilde{T} = T^*$ lies between $T=1$ and $T=\lfloor (\sum_{a_i \in S} a_i) / k \rfloor \le m_1a + m_2b$.
So to find $\tilde{T}$, we need only $O(\log(m_1a + m_2b))$ calls to $A'$ with binary search: any $T$ for which we get a collection of size $\ge k$ from $A'$ is $\le \tilde{T}$; any $T$ for which we get a collection of size $&lt; k$ from $A'$ is $&gt; \tilde{T}$. $\square$
Now we show (or *claim to show*) that $P'$ can be solved in poly-time:

Problem $P'$: given a multiset $S$ of $m_1$ elements equal to $a \in \mathbb{Z}^+$ and $m_2$ elements equal to $b \in \mathbb{Z}^+$, and $T \in \mathbb{Z}^+$, what is the maximally sized disjoint collection of subsets of $S$ which has the property that each subset's sum is $\ge T$?
Claim 2 we can solve $P'$ in poly-time.
Let $\ell^*$ denote the size maximal size disjoint collection (we are looking for $\ell^*$ and a disjoint collection that achieves this maximum).
First some definitions:
Definition call multisets consisting of only $a$'s and $b$'s, &quot;$(a, b)$-multisets&quot;. Any such multiset can be identified by an ordered pair $(i, j)$ where $i$ denotes the number of $a$'s and $j$ denotes the number of $b$'s.
Definition &quot;$(a,b,T)$-minimal multisets&quot; $(i, j)$ are those $(a,b)$-multisets for which removing any element would bring you to a sum below $T$: i.e. $ai + bj \ge T$, but $a(i-1) + bj &lt; T$ and $ai + b(j-1) &lt; T$. For example, if $a=2$ and $b=5$ and $T=20$, the minimal multisets would be $(10, 0)$, $(8, 1)$, $(5, 2)$, $(3, 3)$ and $(0, 4)$.
Observation: Without loss of generality,  we can assume that the optimal collection of $\ell^*$ subsets of $S$ consists of only $(a, b,T)$-minimal multisets. To see why this is without loss of generality, consider any collection of $\ell^*$ subsets of $S$, each with sum $\ge T$: remove as many elements as you can from each of these subsets, while maintaining the invariant that each must have sum $\ge T$. Once you cannot remove any more, you have a collection of $\ell^*$ $(a,b,T)$-minimal subsets. So indeed it is fine to look only for collections of $(a,b,T)$-minimal multisets.
Note that there are not so many &quot;minimal&quot; $(a,b,T)$-multisets that can be subsets of $S$: we can crudely bound by $\min(m_1, m_2)+1$ (since no two minimal multisets can have either coordinate equal).
Call the actual number of these minimal multisets that are also subsets of $S$ as $M$. Order these $M$ subsets in say lexicographically decreasing order (like we did above), and call them $(i_1, j_1), (i_2, j_2), \dots, (i_p, j_p),\dots, (i_M, j_M)$.
We claim to solve $P'$ in poly-time with a dynamic programming approach (where we'll construct the optimal $\ell^*$ minimal multisets in lexicographically decreasing order):
Our overall problem is to compute the maximum number of disjoint $(i_p, j_p)$s we can find among the base set $S = (m_1, m_2)$.
Let $F(M', m_1', m_2')$ be defined as the maximally sized disjoint collection of $(m_1', m_2')$ you can find, where all the subsets must be one of the lexicographically smallest $M'$ subsets of the $M$ $(a,b,T)$-minimal subsets we identified, i.e. $(i_{M-M'+1}, j_{M-M'+1}), (i_{M-M'+2}, j_{M-M'+2}) ..., (i_M, j_M)$.
We want to find $F(M, m_1, m_2)$.
We have the recurrence $$F(M, m_1, m_2) = 1 + \max_{p=1, 2, \dots, M} F(M-p+1, m_1-i_p, m_2-j_p)$$ and similarly $$F(M', m_1', m_2') = 1 + \max_{p=M-M'+1, \dots, M} F(M-p+1, m'_1-i_p, m'_2-j_p).$$ In each subproblem none of the 3 coordinates ever increase, and always either $m_1'$ or $m_2'$ decreases.
The running time of this dynamic programming is bounded by $O(M (Mm_1m_2)) = O((m_1+m_2)^4)$, as we wanted (strictly, the DP only gives the value of $\ell^*$ and not an achieving disjoint $\ell^*$-collection, but I think an actual optimal solution should be able to be constructed from the DP table as well). $\square$.