Find the smallest group of numbers with sum bigger then $X$
Assuming you are looking for $S' \subset S$ with minimal $\sum_{x \in S'} x &gt; X$ like Steven suggested, then you should be able to reduce the problem to an instance of knapsack with weights $w_i$ and costs $c_i$ with $w_i = c_i = s_i$ for every $s_i \in S$ where the weight of the knapsack is bounded by $X$.
If the solution $K := \sum_{x \in S'} x$ is not greater than $X$, you need to repeat the process by setting $X := X + \varepsilon$ where $\varepsilon$ is the minimal distance between any two subset sums of $S$ and running knapsack again until it yields a solution $K &gt; X$.
Edit: Turns out that finding $\varepsilon$ isn't an easy problem. I first assumed $\varepsilon$ is the minimum between any two numbers in $S$, which can be found by sorting the list and finding the minimum between two adjacent elements. This would yield a good approximation in most cases, but it might not be the optimal solution. If your problem statement uses $\geq$ instead of $&gt;$, transforming the problem into knapsack will yield an optimal solution, no repetitions required.