a.
Consider 201 disjoint blocks $B_{1}, B_{2}, \ldots, B_{201}$ consisting each of 10 numbers, consecutive on the circle. By the pigeonhole principle, there exists a block $B_{k}$, for some $1 \leq k \leq 201$, not containing any of the 200 numbers
$$\frac{1}{1 \times 2}, \frac{1}{2 \times 3}, \ldots, \frac{1}{200 \times 201}$$
The sum of the ten consecutive numbers in this block $B_{k}$ is less than or equal to
$$\begin{aligned} \frac{1}{201 \times 202} + & \frac{1}{202 \times 203} + \cdots + \frac{1}{210 \times 211} \\ & = \left(\frac{1}{201} - \frac{1}{202}\right) + \left(\frac{1}{202} - \frac{1}{203}\right) + \cdots + \left(\frac{1}{210} - \frac{1}{211}\right) \\ & = \frac{1}{201} - \frac{1}{211} = \frac{10}{201 \times 211} < \frac{1}{4000} \end{aligned}$$

b.
Again, consider the 201 disjoint blocks $B_{1}, B_{2}, \ldots, B_{201}$ like in (a). By the pigeonhole principle, there exists at least 101 blocks $B_{k_{1}}, B_{k_{2}}, \ldots, B_{k_{201}}$, not containing any of the 100 numbers
$$\frac{1}{1 \times 2}, \frac{1}{2 \times 3}, \ldots, \frac{1}{100 \times 101}.$$
The sum of the 1010 numbers in these 101 blocks $B_{k_{1}}, B_{k_{2}}, \ldots, B_{k_{201}}$ is less than or equal to
$$\begin{aligned} \sum_{i=1}^{201} \left( \sum_{b \in B_{k_{i}}} b \right) & \leq \frac{1}{101 \times 102} + \frac{1}{102 \times 103} + \cdots + \frac{1}{210 \times 211} \\ & \leq \left(\frac{1}{101} - \frac{1}{102}\right) + \left(\frac{1}{102} - \frac{1}{103}\right) + \cdots + \left(\frac{1}{210} - \frac{1}{211}\right) \\ & < \frac{1}{101} \end{aligned}$$
Therefore, there exists at least one block $B_{k_{i_{0}}}$ among these 101 blocks with the sum of its ten consecutive numbers less than
$$\frac{1}{101 \times 101} < \frac{1}{10000}.$$