Order of maximin sets You are allowed to choose between several fractions $p_1/q_1,\ldots,p_n/q_n$ (where $p_i<q_i$ for all $i$). 
Once you select a fraction, say $p/q$, you are shown a finite set $X$ of positive numbers. 
You are allowed to partition $X$ to $q$ subsets. 
Your score is the sum of numbers in the $p$ subsets with the smallest sum. 
The question is: what is the best fraction to select?
Intuitively, larger fraction is better. However, this is not always true. 
For example, suppose $X=\{100,99,98,4,3,2,1\}$. Then:


*

*when $p/q=1/3$, you can partition $X$ to $\{100,1,2\}$, $\{99,3\}$, $\{98,4\}$; the smallest sum is 102.

*when $p/q=2/5$, you can partition $X$ to $\{100\}$, $\{99\}$, $\{98\}$, $\{4,3\}$, $\{2,1\}$; the smallest sum (of the last two subsets) is 10.

*when $p/q=3/7$, the best partition is to individual elements; the smallest sum of three is 6.


I found three cases in which the larger fraction is (weakly) better regardless of $X$:


*

*Same denominator: If $p_1 \geq p_2$, then $p_1/q$ is better than $p_2/q$  for any $q$. Proof: use the optimal partition for $q$ subsets. You get more subsets, so your score is weakly larger.

*Same numerator: If $q_1 \leq q_2$, then $p/q_1$ is better than $p/q_2$ for any $p$. Proof: take the optimal partition to $q_2$, unite some subsets arbitrarily until you remain with $q_1$ subsets. The $p$ sets with the smallest sum has a weakly larger sum than before, so your score is weakly larger.

*Reduced fraction: If $r\geq 1$, then $p/q$ is better than $(p\cdot r) / (q\cdot r)$  for any $p,q$. Proof: take the optimal partition to $q \cdot r$ subsets. Starting from the subsets with the smaller sum, unite each $r$ subsets to a single subset. This results in a partition to $q$ subsets, and the $p$ with the smallest sum are exactly the $r \cdot p$ with the smallest sum in the original partition.
Are there any other cases in which the larger fraction is always better?
 A: Let $p_1 / q_1 \succ p_2 / q_2$ mean that $p_1 / q_1$ is better than $p_2 / q_2$. 
This relation is transitive, so it is possible to combine your rules, adding intermediate fractions. For example $3 / 5 \succ 2 / 5 \succ 2 / 7$ . Generalizing we get that $p_1 / q_1 \succ p_2 / q_2$ for all $p_1 \ge p_2$ and $q_1 \le q_2$.
Another example is $1 / 3 \succ 4 / 12 \succ 4 / 13 \succ 3 / 13$. Generalizeing we get that $p_1 / q_1 \succ p_2 / q_2$ for all $p_1, p_2, q_1, q_2$ such that
$$q_2 \ge q_1 \text{ and } p_1 \cdot \left\lfloor \frac{q_2}{q_1}\right\rfloor \ge p_2. \tag{1}$$
Proof.
$$p_1 / q_1 \succ \left(\left\lfloor \frac{q_2}{q_1}\right\rfloor p_1\right) \Big/ \left(\left\lfloor \frac{q_2}{q_1}\right\rfloor q_1\right) \succ p_2 / q_2.$$
Note that condition (1) is true in all your three cases.

Another sufficient condition is $q_1 \ge q_2$ and $q_1 - p_1 \le q_2 - p_2$. 
For example $9 / 10 \succ 7 / 9$.
Proof. Consider partition of $X$ into $q_2$ parts. Add $q_1 - q_2$ empty sets. Taking $q_1 - q_2$ empty sets and $p_2 \le p_1 - (q_1 - q_2)$ more we get for the first fraction the same sum as for the second fraction and we have to add $p_1 - p_2 - q_1 + q_2 \ge 0$ subsets.  

Condition $p_1 / q_1 \ge p_2 / q_2$ is necessary for $p_1 / q_1 \succ p_2 / q_2$.
Proof. Let $p_1 / q_1 < p_2 / q_2$. Consider $X = \{\,x, x + 1, \ldots, x + q_1q_2 - 1\,\}$ for sufficiently large $x$. Then for the first fraction we get less than $p_1q_2(x + q_1q_2)$ and for the second fraction at least $p_2q_1x.$ So for $x > p_1q_2q_1q_2 / (p_2q_1 - p_1q_2)$ the second fraction becomes better. So $p_1 / q_1 \nsucc p_2 / q_2$ in this case.

Condition $q_1 \le q_2$ or $q_1 - p_1 \le q_2 - p_2$ is necessary for $p_1 / q_1 \succ p_2 / q_2$.
Proof. Let $q_1 > q_2$ and $q_1 - p_1 > q_2 - p_2$. Consider $X = \{\,1, 2, \ldots, q_1 - q_2; x, x + 1, \ldots, x + q_2 - 1\,\}$ for sufficiently large $x$. Then for the first fraction we get less than $\frac{(q_1 - q_2)(q_1 - q_2 + 1)}{2} + \max\{\,0, p_1 - q_1 + q_2\,\}(x + q_2)$. For the second fraction we get at least $p_2 x$. For $x > \frac{(q_1 - q_2)(q_1 - q_2 + 1)}{2}$ we get that second fraction is better. So $p_1 / q_1 \nsucc p_2 / q_2$ in this case.

So the only remaining case is $q_1 \le q_2$ and $p_1 / q_1 \ge p_2 / q_2$ and $p_1 \cdot \left\lfloor \frac{q_2}{q_1} \right\rfloor < p_2$. I guess that such fractions are incomparable, i. e. for each such pair of fractions there are sets $X_1$ and $X_2$ such that for $X_1$ the first fraction is better, while for $X_2$ the second fraction is better.
