If $k \geq 14$, then there are at least $1+2+\ldots+14=105>100$ stones, contradiction. So $k \leq 13$. We can take $\{1, 2, \ldots, 11, 12, 22 \}$ for $k=13$; it's easy to check that this works. So the maximum $k$ is $13$. Note that $k=10$ works, taking $\{1, 3, 5, \ldots, 19 \}$: there are $100$ stones, and if we partition a pile, we get an odd number and an even number; the odd number is among the smaller piles. Now we want to show that $k<10$ doesn't work. Suppose that there's no 1-stone pile. Then, since $n=(n-1)+1$, if there's a pile with $n$ stones there must be one with $n-1$. So, if $n$ is number of stones in the biggest pile, there are $n+(n-1)+\ldots+2=(n+1)n/2-1$ stones, which can't be $100$. Then there's a pile with just $1$ stone. Suppose that the biggest pile has $2m+1$ stones. Then at least one of $1, 2m$, one of $2, 2m-1$, ..., one of $m, m+1$ must be present (*), so at least $m+1$ piles in total. If the biggest pile has $2m$ stones, there are at least $m$ piles. Note that (*) can be applied to any pile, in fact. Now, suppose $k \leq 8$. Then the biggest pile has at most $2k=16$ stones. So there are at most $1+16+15+\ldots+10=92<100$ stones, contradiction. Then $k \geq 9$. If $k=9$, the biggest pile has at most $18$ stones. Take the smallest pile $>10$; by (*), there are at least 5 piles $\leq 10$. If the biggest pile has $17$ or fewer stones, there are at most $1+7+8+9+10+14+15+16+17=97<100$ stones, so it has $18$. However, the smallest pile $>1$ can be at most $4$ by (*), so in any case there are at most $1+4+8+9+10+15+16+17+18=98$ stones. Then $k>9$.

During the Olympiad I could prove that $k_m = 13$ and $k \geq9$ and I got 2 points out of 7

A shorter and nicer(maybe not) solution for the minium part: Let $a_{1}<a_{2}<a_{3}\dots<a_{n}$$a_{1}+a_{2}+\dots+a_{k}=100$ we can prove that $a_{k}\leq 2k$ if $a_{k}>2k$ then consider the pairs of number $(1,a_{k}-1),(2,a_{k}-2)\dots(k,a_{k}-k)$ .there must exist $j$ such that $j\not =a_{i},a_{k}-j\not =a_{i}$for any $i$.this is a contradution. then we have $a_{1}+\dots+a_{9}\leq 90$ so $k\geq 10$

consider the smallest pile with l stones if $l>2$,then partition it into 1 and $l-1$,contradiction so $l=1 or 2$ if $l=2$,then the second small pile (for the same reason) contains 3 stones,then 4 stones,5 stones,... but 2+3+4+...+n cannot equal to 100,contradiction hence $l=1$ let the largest pile has m stones if $k\le 9$then $m+(m-1)+...+(m-7)+1\ge 100$ hence $m\ge 19$ then m has at least 9 methods to be partitioned,with 8 other piles remaining,so there exists a method which contradicts the condition given. so $k\ge 10$ and 1,3,5,...,19 give an example for 10.

IF stones : 1+2+ .. + n=100 $implies n is not \in \mathbb (Z)$ If we find for max :1 + 2 + 3+.... + n + 2n =100 or 1 + 2 + 3+.... + n + 2n-1=100 $\implies$ we find max 13 : 1 + 2 +3 + .... + 12 + 22 =100 we have this : we get #$n$ so we put after #$n$ , $n+1$ or $2n$ ($n$ is even) ($2n + 1$ , $n$ is odd) (#$n = 1,2,...,n $) $\implies $ max = 13 $O.Y.SH.$

If $k=1$ then it is clearly impossible. So $k\ge2$. If the smallest pile has at least $3$ stones, we can split it into two piles with $1$ and $2$ stones, so the smallest pile has at most $2$ stones. If it has two stones, then, if there are $x$ stones in another pile, there must be $x-1$ stones in another pile by splitting it into piles with $1$ and $x-1$ stones. Then there are $\sum_{i=2}^{k+1}i=\frac{(k+1)(k+2)}2-1$ stones, and solving for $k$ gives $k=\frac{-3+\sqrt{805}}2$ which isn't an integer. So the smallest pile has exactly one stone. Then if the second-smallest pile is at least $4$ in size, we can split it into two piles with at least two stones, so the second-smallest pile can be at most $3$ in size. If the largest pile has $2k+1$ stones, then there are $k$ partitions $(1,2k),(2,2k-1),\ldots,(k,k+1)$, which each correspond to at least one pile of stones. But the largest pile isn't any one of these piles, so there are at least $k+1$ piles, contradiction. So the largest pile has at most $2k$ stones. We can claim now that the minimum and maximum of $k$ are $10$ and $13$, respectively. Suppose $k\ge14$. Then we have at least $\sum_{i=1}^{14}i=105>100$ stones, so there cannot be a good partition. On the other hand, suppose $k\le9$. Then there are at most $1+3+\sum_{i=10}^{16}i=95<100$ stones, contradiction. For $k=10$ we can take piles with sizes $\{1,3,5,\ldots,19\}$ whereas for $k=13$ we can take piles with sizes $\{1,2,3,\ldots,12,22\}$. Remark: $k=11$ and $k=12$ work too.

littletush wrote: ... if $k\le 9$then $m+(m-1)+...+(m-7)+1\ge 100$ hence $m\ge 19$ ....... We get $m \ge 17$.

jasperE3 wrote: ...... Then if the second-smallest pile is at least $4$ in size, we can split it into two piles with at least two stones, so the second-smallest pile can be at most $3$ in size. ....... On the other hand, suppose $k\le9$. Then there are at most $=1+3+\sum_{i=10}^{16}i=95<100$ stones, contradiction. ...... How does $k\le9$ imply $\sum_{i=1}^{k}a_i \le 1+3+\sum_{i=10}^{16}i=95<100$? What if $a_k \ge 17$, for example?

Hawk Tiger wrote: A shorter and nicer(maybe not) solution for the minium part: Let $a_{1}<a_{2}<a_{3}\dots<a_{n}$$a_{1}+a_{2}+\dots+a_{k}=100$ we can prove that $a_{k}\leq 2k$ if $a_{k}>2k$ then consider the pairs of number $(1,a_{k}-1),(2,a_{k}-2)\dots(k,a_{k}-k)$ .there must exist $j$ such that $j\not =a_{i},a_{k}-j\not =a_{i}$for any $i$.this is a contradution. then we have $a_{1}+\dots+a_{9}\leq 90$ so $k\geq 10$ We have only $a_k \le 2k$, where $k$ is the number of stones in the pile with the largest amount of stones. For the rest $i$'s we do not know if $a_i \le 2i$, do we?!