For every $m$ and $k$ integers with $k$ odd, denote by $\left[\frac{m}{k}\right]$ the integer closest to $\frac{m}{k}$. For every odd integer $k$, let $P(k)$ be the probability that
\[\left[\frac{n}{k}\right] + \left[\frac{100 - n}{k}\right] = \left[\frac{100}{k}\right]\]
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 $1 \leq k \leq 99$?
$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{50}{99} \qquad \textbf{(C)}\ \frac{44}{87} \qquad \textbf{(D)}\  \frac{34}{67} \qquad \textbf{(E)}\  \frac{7}{13}$
Solution 1
Answer: $(D)  \frac{34}{67}$

First of all, you have to realize that 
if $\left[\frac{n}{k}\right] + \left[\frac{100 - n}{k}\right] = \left[\frac{100}{k}\right]$
then $\left[\frac{n - k}{k}\right] + \left[\frac{100 - (n - k)}{k}\right] = \left[\frac{100}{k}\right]$
So, we can consider what happen in $1\le n \le k$ and it will repeat. Also since range of $n$ is $1$ to $99!$, it is always a multiple of $k$. So we can just consider $P(k)$ for $1\le n \le k$.

Let $\text{fpart}(x)$ be the fractional part function
This is an AMC exam, so use the given choices wisely. With the given choices, and the previous explanation, we only need to consider $k = 99$, $87$, $67$, $13$. $1\le n \le k$

For $k > \frac{200}{3}$, $\left[\frac{100}{k}\right] = 1$. 3 of the $k$ that we should consider land in here.
For $n  \frac{k}{2}$ .  ( $k > \frac{200}{3}$, no worry for the rounding to be $> 1$)
$100 > k > \frac{k}{2} + n$, so this is always true.
For $\frac{k}{2} n > 100 - \frac{67}{2} = 66.5$
For k = 69, $69 > n > 100 - \frac{69}{2} = 67.5$
etc.

We can clearly see that for this case, $k = 67$ has the minimum $P(k)$, which is $\frac{34}{67}$. Also, $\frac{7}{13} > \frac{34}{67}$ .
So for AMC purpose, answer is $\boxed{\textbf{(D) }\frac{34}{67}}$.

Proof:
Notice that for these integers $99,87,67$:

$0\rightarrow 49,50,51\rightarrow 98$
$100\rightarrow 51,50,49\rightarrow 2$
$P=\frac{98}{99}$

$0\rightarrow 43,44\rightarrow 56,57\rightarrow 86$
$87\rightarrow 57,56\rightarrow 44,43\rightarrow 14$
$P=\frac{74}{87}$

$0\rightarrow 33,34\rightarrow 66$
$100\rightarrow 67,66\rightarrow 34$
$P=\frac{34}{67}$

That the probability is $\frac{2k-100}{k}=2-\frac{100}{k}$. Even for $k=13$, $P(13)=\frac{9}{13}=\frac{100}{13}-7$. And $P(11)=\frac{10}{11}=10-\frac{100}{11}$.
Perhaps the probability for a given $k$ is $\left\lceil{\frac{100}{k}}\right\rceil-\frac{100}{k}$ if $\left[\frac{100}{k}\right]=\left\lfloor{\frac{100}{k}}\right\rfloor$ and $\frac{100}{k}-\left\lfloor{\frac{100}{k}}\right\rfloor$ if $\left[\frac{100}{k}\right]=\left\lceil{\frac{100}{k}}\right\rceil$.
So $P>\frac{1}{2}$ and $P_\text{min}=\frac{k_\text{min}+1}{2k_\text{min}}=\frac{101}{201}$. Because $201=3\cdot 67\mid 99!$  !


Now, let's say we are not given any answer, we need to consider $k  \frac{1}{2}$ and $\text{fpart} \left(\frac{n}{k}\right) > \frac{1}{2}$
Case 1)
$\text{fpart} \left(\frac{100 - n}{k}\right)  $\text{fpart}\left(\frac{100}{k}\right) = \text{fpart} \left(\frac{n}{k}\right) +\text{fpart} \left(\frac{100 - n}{k}\right)$
Case 2)
$\text{fpart} \left(\frac{100 - n}{k}\right) > \frac{1}{2}$
-> $\text{fpart}\left(\frac{100}{k}\right) + 1 = \text{fpart} \left(\frac{n}{k}\right) +\text{fpart} \left(\frac{100 - n}{k}\right)$

and for $n = 1$, since $k$ is odd, $\left[\frac{99}{k}\right] \neq \left[\frac{100}{k}\right]$
-> $99.5 = k (p + .5)$  -> $199 = k (2p + 1)$, and $199$ is prime so $k = 1$ or $k =199$, which is not in this set
, which makes 
$P(k) \ge \frac{1}{2} + \frac{1}{2k}$.

Now the only case without rounding, $k = 1$. It must be true.

Solution 2
It suffices to consider $0\le n 67$ $87$ and $99$ do not yield better probabilities. Therefore, our answer is $\boxed{\frac{34}{67}}.$