The least number of special coins is 101. Example with exactly 101 special coins: The first division has 200 groups with 101 coins each; the second division is obtained by dividing one of these groups into 101 groups of 1 coin.
Let $x_1 \le x_2 \le \dots \le x_{200}$ be the sizes of the 200 groups in the first division. Suppose that the second division has 200 groups without any special coin (there may be more such groups), and let their sizes be $y_1 \le y_2 \le \dots \le y_{200}$. Clearly $x_1 + x_2 + \dots + x_{200} > y_1 + y_2 + \dots + y_{200}$ since the second division has more than 200 groups. Hence there is an index $j = 1, \dots, 200$ such that $x_j > y_j$. Assume $j$ to be minimal with this property, meaning that $x_i \le y_1, \dots, x_{j-1} \le y_{j-1}, x_j > y_j$.
Consider a group $y_i$ with $1 \le i \le j$. Each coin in it is not special, so in the first division it was in a group of size $\le y_i$. On the other hand $x_{200} \ge \dots \ge x_j > y_j \ge y_i$, hence groups of size $\le y_i$ in the first division are among $x_1, \dots, x_{j-1}$. Thus the entire group $y_i$ is contained in the union of $x_1, \dots, x_{j-1}$. The conclusion holds for every $i=1, \dots, j$, so the union of $y_1, \dots, y_j$ is contained in the union of $x_1, \dots, x_{j-1}$. However this is false as $x_i \le y_1, \dots, x_{j-1} \le y_{j-1}$ and $y_j > 0$.
Thus the second division has at most 199 groups without any special coin. Then there are at least 101 groups with a special coin in it, yielding at least 101 special coins in particular.