Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $B$ at a speed of $7$ meters per second and follows the straight path that produces the earliest possible meeting of the two skaters, given their speeds. How many meters does Allie skate before meeting Billie?

[asy]
defaultpen(linewidth(0.8));
draw((100,0)--origin--60*dir(60), EndArrow(5));
label("$A$", origin, SW);
label("$B$", (100,0), SE);
label("$100$", (50,0), S);
label("$60^\circ$", (15,0), N);[/asy]
1. **Define the problem and variables:**
   - Let \( A \) and \( B \) be the initial positions of Allie and Billie, respectively.
   - The distance between \( A \) and \( B \) is \( AB = 100 \) meters.
   - Allie skates at a speed of \( 8 \) meters per second along a path that makes a \( 60^\circ \) angle with \( AB \).
   - Billie skates at a speed of \( 7 \) meters per second along a path that will allow them to meet Allie as soon as possible.

2. **Determine the meeting point \( P \):**
   - Let \( AP = x \) be the distance Allie skates before meeting Billie.
   - Since Allie skates at \( 8 \) meters per second, the time \( t \) it takes for Allie to reach \( P \) is \( t = \frac{x}{8} \).

3. **Express Billie's path and time:**
   - Let \( BP = y \) be the distance Billie skates before meeting Allie.
   - Since Billie skates at \( 7 \) meters per second, the time \( t \) it takes for Billie to reach \( P \) is \( t = \frac{y}{7} \).

4. **Equate the times:**
   \[
   \frac{x}{8} = \frac{y}{7}
   \]
   \[
   y = \frac{7}{8}x
   \]

5. **Apply the Law of Cosines to triangle \( ABP \):**
   - The angle \( \angle PAB = 60^\circ \).
   - Using the Law of Cosines:
   \[
   y^2 = x^2 + 100^2 - 2 \cdot x \cdot 100 \cdot \cos(60^\circ)
   \]
   \[
   y^2 = x^2 + 10000 - 100x
   \]
   - Substitute \( y = \frac{7}{8}x \):
   \[
   \left( \frac{7}{8}x \right)^2 = x^2 + 10000 - 100x
   \]
   \[
   \frac{49}{64}x^2 = x^2 + 10000 - 100x
   \]

6. **Solve for \( x \):**
   \[
   \frac{49}{64}x^2 = x^2 + 10000 - 100x
   \]
   \[
   49x^2 = 64x^2 + 640000 - 6400x
   \]
   \[
   0 = 15x^2 - 6400x + 640000
   \]
   \[
   15x^2 - 6400x + 640000 = 0
   \]
   - Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   a = 15, \quad b = -6400, \quad c = 640000
   \]
   \[
   x = \frac{6400 \pm \sqrt{6400^2 - 4 \cdot 15 \cdot 640000}}{2 \cdot 15}
   \]
   \[
   x = \frac{6400 \pm \sqrt{40960000 - 38400000}}{30}
   \]
   \[
   x = \frac{6400 \pm \sqrt{2560000}}{30}
   \]
   \[
   x = \frac{6400 \pm 1600}{30}
   \]
   \[
   x = \frac{8000}{30} \quad \text{or} \quad x = \frac{4800}{30}
   \]
   \[
   x = \frac{800}{3} \quad \text{or} \quad x = 160
   \]

7. **Select the smallest positive solution:**
   - The smallest positive solution is \( x = 160 \).

\(\blacksquare\)

The final answer is \( \boxed{ 160 } \) meters.