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

Then $S_j - S_i \doteq 100$, and, on the other hand, $S_j - S_i < S_{100} + a_{201} = 200$. It follows that $S_j - S_i = 100$. Therefore, $a_{i+1} + \dots + a_j = 100$, and the corresponding coins combine into the second group. The remaining coins combine into the third group.

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 equaled $l$. Otherwise, there exists an $j$, 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 equaled $l$.