Belarusian Mathematical Olympiad

Answer: $201$.

Let $a_1\le a_2\le \cdots \le a_n$ be the weights of the coins in Bob's collection. If there exists an $i$ such that $a_i\ge 101$, and $k=l=m=100$, then, obviously, Bob cannot partition his collection into three equally weighted groups. It is easy to see that if $n\le 200$, then there can exist an $a_i\ge 101$ (e.g., $a_1=\cdots =a_{n-1}=1$, $a_n=300-(n-1)\ge 300-200=100$). Therefore, $n\ge 201$.

We now show that the smallest possible value of $n$ is $201$.

First, note that if $n=201$, then $a_1=\cdots =a_i=1$, where $i\ge 102$. Indeed, if $i\le 101$, then $2\le a_{i+1}\le \cdots \le a_n$, and in this case $a_1+\cdots +a_n\ge 1\cdot i+2\cdot (201-i)=402-i\ge 402-101=301$, contrary to the problem condition.

Without loss of generality, we suppose that $k\le l\le m$. Thus $k\le 100$. So the first $k$ coins combine into the first group ($a_1+\cdots +a_k=k$).

We renumber the remaining $201-k$ weights: $1=x_1=x_2=\cdots =x_t<x_{t+1}\le \cdots \le x_{201-k}$. The total weight of these coins is equal to $$300-k\le x_{201-k}+(201-k-1-t)\cdot 2+t$$ which implies $$t\ge x_{201-k}+100-k.$$ In particular, $t\ge x_{201-k}$ since $k\le 100$.

Now we show how to form the third group with the weight equal to $m$.

Consider the sums: $$\begin{gathered} S_1=a_{201-k},S_2=a_{201-k}+a_{201-k-1},\dots , \\ {S}_i=a_{201-k}+a_{201-k-1}+\cdots +a_{201-k-i+1},i=1,2,\dots \end{gathered}$$ If one of these sums is equal to $m$, then the corresponding coins combine into the third group, and the remaining coins combine into the second group with the weight equal to $l$. Otherwise, there exists a $j$, such that $S_j<m$ and $S_{j+1}>m$. In particular, $S_{j+1}-S_j\ge 2$, i.e. $x_{201-k-j}\ge 2$. This means that among the numbers $x_{201-k}$, $x_{201-k-1}$, ..., $x_{201-k-j}$ there are no $1$'s. Moreover, this means that $m-S_j<x_{201-k-j}\le x_{201-k}\le t$. So $$x_1+\cdots +x_{m-S_j}+x_{201-k-j+1}+\cdots +x_{201-k}=m,$$ and the corresponding coins combine into the third group. The remaining coins combine into the second group with weight equal to $l$.