Does every allocation have a maximal Pareto-improvement? Consider an economy with a finite number of goods and a finite quantity of each good. Each agent $i$ has a preference-relation $\succeq_i$ which is a total, reflexive and transitive relation over the set of bundles. Given an allocation $x$ of the goods among the agents, say that $x^*$ is a maximal Pareto-improvement of $x$ if:


$x^*$ is Pareto-optimal,
Every agent weakly prefers the allocation $x^*$ over the allocation $x$.


What is the weakest condition on the preference-relations $\succeq_i$ that guarantees that every allocation has a maximal Pareto-improvement?

I would guess that it is somehow related to compactness, however, I do not see how to define this condition on preference-relations.
I think there is a short proof if you also assume that the number of agents $n$ is finite and that the preferences are continuous.

Given the second assumption Debreu's theorem (1954, "Representation of a preference ordering by a numerical function") states that a continuous utility function exists which represents the preferences. I will denote the utility function representing the preference of agent $i$ by $U_i$.

Though this is somewhat uncommon, I will denote the utility agent $i$ gets from an allocation $x$ by $U_i(x)$.

An algorithm resulting in a maximum Pareto-improvement of $x$:
Step 1. Initially set $i=1$ and set $y = x$.
Step 2. Maximize the utility of agent $i$ over the set of allocation where all other agents $j$ have at least utility $U_j(y)$. As all functions $U_j$ are continuous, there is a finite number of goods and a finite quantity of each good, for all $j$ the upper set of $U_j$ is a closed set. A finite intersection of closed sets is also a closed set. As $U_i$ is also continuous, it will have a maximum over this set. Select an allocation $y'$ which maximizes $U_i$ and let this be the our allocation $y$ from now on, so set $y = y'$.
Step 3. If $i = n$, stop. Otherwise set $i = i+1$ and proceed to step 2.

Notice the following:


Any move from $y$ to $y'$ in Step 2. is a Pareto-improvement.
Pareto-improvements are transitive, therefore the set of Pareto-improvements is narrowing, i.e. if $y'$ is a Pareto-improvement to $y$ then the set of Pareto-improvements to $y'$ is included in the set of Pareto-improvements to $y$.
As the utility of agent $i$ is maximized in Step 2. over the set of Pareto-improvements to $y$, it is impossible that his utility can be further increased by a Pareto-improvement. Therefore once we reach Step 3. for $i = n$ we have a Pareto-optimal allocation $y$, that is either a Pareto-improvement of $x$ or where $\forall i$: $x\sim_i y$. Hence $y$ is a maximal Pareto-improvement of $x$.




A comment: It seems to me you could replace continuity with monotonicity, but that would require another proof. Hopefully someone else is up to it!