I have a conjecture: Let x be the minimum natural number such that: x + (x+1) + (x+2).... + (x+n-1)>m Then let S={x,x+1...k} Notice that it has the desirable (but inconclusive) property that if I add in another element to this set, it no longer satisfies the condition. It can be proven easily for n=2, and probably not too hard for n=3 either. (n=1 it is not true). Any ideas? very curious about this.

Someone reminded me of this problem and so I have given it another shot Let $x(n,m,k)$ be the number $x$ as given in the above post. Let $f(n,m,k)$ be the size of the set $S$ in the above post. Notice: $x(n,m,k)<x(n',m',k) <=> f(n,m,k)>f(n',m',k)$ Now suppose $N>2$ we make the inductive hypothesis that $f(n,m,k)$ is indeed the answer to the problem for $n=2,3...N-1$ (the base case $n=2$ is easy and I will omit it). Suppose we have chosen $f(N,m,k)+1$ numbers. Then we have some number $y$ that is less than $x(N,m,k)$. Let $m'=m-y$. Now remove $y$ from our set. We have $f(N,m,k)$ numbers in our set. Now we want to show that $f(N,m,k)>f(N-1,m',k)$ and we will do this if we can show that $x(N,m,k)<x(N-1,m',k)$. Let $x=x(N,m,k)$ then we have: $(x-1)+x+(x+1)+...+(x+N-2)\leq m$ And so: $x+(x+1)...+(x+N-2) \leq m-(x-1) \leq m-y=m'$ (since $y<x$ as mentioned above) And so this shows that $x(N-1,m',k)>x(N,m,k)$ Hence $f(N-1,m',k)<f(N,m,k)$. Now we use the induction hypothesis: We have $f(N,m,k)$ numbers in our set hence we have more than $f(N-1,m',k)$ numbers in our set. That means by the induction hypothesis that we can choose $N-1$ of them such that their sum is $m'$, but now we can add in the number $y$ to this set and the sum is exactly $m$. So in summary $f(n,m,k)$ is the answer to the problem for $n>1$, while for $n=1$ it is $k-1$ (choose every number other than $m$) This number is easy to compute hence I declare it a suitable solution to the problem