If we divide $200 \cdot 101$ coins firstly in $200$ groups with $101$ coins each, and then one of these groups we divide into $101$ groups of one coin each, we would obtain $300$ groups with exactly $101$ special coins. In general let $x_1 \leq x_2 \leq \ldots \leq x_{200}$ be the amounts of coins in the first grouping. If we prove that in the second grouping there are at most $199$ groups without a special coin, then we would have at least $300 - 199 = 101$ groups with at least one special coin in each and thus the special coins would be at least $101$. Suppose otherwise, i.e. that there exist $200$ groups without special coins and let $y_1 \leq y_2 \leq \ldots \leq y_{200}$ be the amounts of coins in these groups. Since $\sum_{i=1}^{200} x_i > \sum_{i=1}^{200} y_i$, there exists a $j$, such that $x_i \leq y_i$ for all $i\leq j-1$ and $x_j > y_j$. The coins in each groups $y_i$, $i \leq j-1$, are not special, so in the first grouping they have been in a group with amount of coins, not exceeding $y_i$. But all groups $x_{j+1}, x_{j+2}, \ldots, x_{200}$ have more coins than all of $y_i$, $i \leq j-1$. Hence all coins in $y_1,\ldots,y_j$ are among the coins $x_1,x_2,\ldots,x_{j-1}$. Therefore $\sum_{k=1}^j y_k \leq \sum_{k=1}^{j-1} x_k$, contradicting $x_i \leq y_i$ for $i\leq j-1$ and $y_j > 0$. The result folows.