Answer: $201$.

Let $a_1 \le a_2 \le \dots \le a_n$ 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., $a_1 = \dots = a_{n-1} = 1$, $a_n = 300 - (n-1) \ge 300 - 200 = 100$). Therefore, $n \ge 201$.

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

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

We renumber the remaining $201-k$ weights: $1 = x_1 = x_2 = \dots = x_t < x_{t+1} \le \dots \le x_{201-k}$. 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:
$$S_1 = a_{201-k}, \quad S_2 = a_{201-k} + a_{201-k-1}, \quad \dots, \\ S_i = a_{201-k} + a_{201-k-1} + \dots + a_{201-k-i+1}, \quad i = 1, 2, \dots$$
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
$$x_1 + \dots + x_{m-S_j} + x_{201-k-j+1} + \dots + x_{201-k} = m,$$
and the corresponding coins combine into the third group. The remaining coins combine into the second group with weight equal to $l$.