For every m and k integers with k odd, denote by \left[\dfrac{m}{k}\right] the integer closest to \dfrac{m}{k}. For every odd integer k, let P(k) be the probability that
\left[\dfrac{n}{k}\right]+\left[\dfrac{100-n}{k}\right]=\left[\dfrac{100}{k}\right]
for an integer n randomly chosen from the interval 1 \leq n \leq 99 !. What is the minimum possible value of P(k) over the odd integers k in the interval 1 \leq k \leq 99 ?
Answer Choices
A. \dfrac{1}{2}
B. \dfrac{50}{99}
C. \dfrac{44}{87}
D. \dfrac{34}{67}
E. \dfrac{7}{13}