AIME II

Let $x_1 \le x_2 \le \dots \le x_{100}$, $S = x_1 + x_2 + \dots + x_{100} = 0$, and $T = |x_1| + |x_2| + \dots + |x_{100}| = 1$.

We want to maximize $x_{76} - x_{16}$.

Let $k$ be the number of negative $x_i$'s, and $100-k$ the number of nonnegative $x_i$'s. Since the sum is $0$, the sum of the negative numbers equals the sum of the positive numbers in magnitude.

Let $x_1 = x_2 = \dots = x_k = -a$ and $x_{k+1} = x_{k+2} = \dots = x_{100} = b$, with $a, b \ge 0$.

Then:

$k(-a) + (100-k)b = 0 \implies (100-k)b = ka \implies b = \frac{k}{100-k}a$

Also,

$k|{-a}| + (100-k)|b| = ka + (100-k)b = 1$

Substitute $b$:

$ka + (100-k)\left(\frac{k}{100-k}a\right) = 1$

$ka + k a = 1$

$2ka = 1 \implies a = \frac{1}{2k}$

Then $b = \frac{k}{100-k}a = \frac{k}{100-k} \cdot \frac{1}{2k} = \frac{1}{2(100-k)}$

Now, $x_{76} - x_{16}$:

Since $x_1 = \dots = x_k = -a$, $x_{k+1} = \dots = x_{100} = b$.

The first $k$ entries are $-a$, the next $100-k$ are $b$.

$x_{16}$ is $-a$ if $k \ge 16$, otherwise $b$.

$x_{76}$ is $-a$ if $k \ge 76$, otherwise $b$.

We want to maximize $x_{76} - x_{16}$.

Case 1: $k < 16$

Then $x_{16} = b$, $x_{76} = b$, so $x_{76} - x_{16} = 0$.

Case 2: $16 \le k < 76$

Then $x_{16} = -a$, $x_{76} = b$, so $x_{76} - x_{16} = b - (-a) = a + b$

Case 3: $k \ge 76$

Then $x_{16} = -a$, $x_{76} = -a$, so $x_{76} - x_{16} = 0$

So, the maximum occurs for $16 \le k < 76$, and $x_{76} - x_{16} = a + b = \frac{1}{2k} + \frac{1}{2(100-k)}$

Let $f(k) = \frac{1}{2k} + \frac{1}{2(100-k)}$ for $16 \le k < 76$.

We want to maximize $f(k)$.

Let’s write $f(k) = \frac{100 - k + k}{2k(100 - k)} = \frac{100}{2k(100 - k)}$

So $f(k)$ is maximized when $k(100 - k)$ is minimized for $16 \le k < 76$.

$k(100 - k)$ is minimized at the endpoints, i.e., $k = 16$ or $k = 75$.

For $k = 16$:

$f(16) = \frac{100}{2 \cdot 16 \cdot 84} = \frac{100}{2 \cdot 1344} = \frac{100}{2688} = \frac{25}{672}$

For $k = 75$:

$f(75) = \frac{100}{2 \cdot 75 \cdot 25} = \frac{100}{2 \cdot 1875} = \frac{100}{3750} = \frac{2}{75}$

Now, $\frac{25}{672} \approx 0.0372$, $\frac{2}{75} \approx 0.0267$, so $\frac{25}{672}$ is larger.

Therefore, the maximum is $\frac{25}{672}$, so $m = 25$, $n = 672$, $m+n = 697$.

Answer: $697$.