a) Let $p_i$ be the $i$-th prime number. Consider the numbers
$$p_1,\ p_1^2 p_2,\ p_1^2 p_2^2 p_3,\ \dots,\ p_1^2 p_2^2 \dots p_{2009} p_{2010}.$$
If $p_1^2 p_2^2 \dots p_k^2 p_{k+1}$ is the least number in the sum, then the sum is divisible by $p_{k+1}$, but not divisible by $p_{k+1}^2$. Hence each sum formed from the given numbers is not a power number.

b) Let us show that for any given integers $a_1, a_2, \dots, a_n$ there exists $b \in \mathbb{N}$ such that $b a_1, b a_2, \dots, b a_n$ are all power numbers. Suppose that $a_i = p_1^{\alpha_{i1}} p_2^{\alpha_{i2}} \dots p_k^{\alpha_{ik}}$, $i = 1, n$, $0 \le \alpha_{ij}$ and $b = p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k}$. If $b a_i$ is a power number, then there exists $q_i$ ($q_i > 1$) such that $\alpha_1 + \alpha_{i_1}, \alpha_2 + \alpha_{i_2}, \dots, \alpha_k + \alpha_{i_k}$ are all divisible by $q_i$.

Let $q_i$ be the $i$-th prime number. By the Chinese Remainder Theorem there exists $\alpha_s$ such that $\alpha_s \equiv -\alpha_{i_s} \pmod{q_s}$, for $i = 1, n$ and $s = 1, k$.

Now we choose arbitrary $2010$ positive integers: $a_1, a_2, \dots, a_{2010}$. Let $S_1, S_2, \dots, S_{2010-1}$ be all the sums formed by them. Then there exists $b$ such that $b S_1, b S_2, \dots, b S_{2010-1}$ are all power numbers. Thus, $b a_1, b a_2, \dots, b a_{2010}$ are the desired numbers.