1. Assigning Weights to Cows:
Each cow at position $n$ is assigned a weight $h^n$, where $h = 1.000000159 = 1 + \frac{159}{10^9}$. When a cow moves from position $m$ to $n$, its weight changes from $h^m$ to $h^n$.

2. Weight Change Analysis:
We need to show that moving the cows strictly decreases their total weight. Consider an interval $[0, 5000]$ without loss of generality. Suppose there are 2501 cows initially at positions $0, 1, \ldots, 2500$. After moving, they occupy positions $2501, 2502, \ldots, 5000$.

3. Initial and Final Weights:
- Initial weight: $\sum_{i=0}^{2500} h^i = \frac{h^{2501} - 1}{h - 1}$
- Final weight: $\sum_{i=2501}^{5000} h^i = \frac{h^{5001} - 1}{h - 1} - \frac{h^{2501} - 1}{h - 1}$

4. Weight Decrease Condition:
If the initial weight is greater than the final weight, we must have:
$$h^{5001} - 2h^{2501} + 1 < 0$$

5. **Approximating $h^{5001}$:**
Using the binomial expansion:
$$h^{5001} = \left(1 + \frac{159}{10^9}\right)^{5001} \approx 1 + 5001 \cdot \frac{159}{10^9} + \binom{5001}{2} \left(\frac{159}{10^9}\right)^2 + \binom{5001}{3} \left(\frac{159}{10^9}\right)^3 + \cdots$$
We can show that higher-order terms are negligible.

6. Simplifying the Expansion:
- First term: $1$
- Second term: $5001 \cdot \frac{159}{10^9} = 0.000795159$
- Third term: $\binom{5001}{2} \left(\frac{159}{10^9}\right)^2 = \frac{5001 \cdot 2500 \cdot 159^2}{10^{18}} = 0.0000003160757025$
- Fourth term: $\binom{5001}{3} \left(\frac{159}{10^9}\right)^3 \approx 0.0000000001024$
- Fifth term and beyond are even smaller.

Summing these, we get:
$$h^{5001} < 1.0007954754281025$$

7. **Approximating $h^{2501}$:**
Similarly,
$$h^{2501} = \left(1 + \frac{159}{10^9}\right)^{2501} \approx 1 + 2501 \cdot \frac{159}{10^9} + \binom{2501}{2} \left(\frac{159}{10^9}\right)^2 + \cdots$$
- Second term: $2501 \cdot \frac{159}{10^9} = 0.000397659$
- Third term: $\binom{2501}{2} \left(\frac{159}{10^9}\right)^2 = 0.00000007903472625$

Summing these, we get:
$$h^{2501} > 1.00039773803472625$$

8. Combining Inequalities:
$$h^{5001} < 1.0007954754281025$$
$$-2h^{2501} < -2.0007954760694525$$
Adding these:
$$h^{5001} - 2h^{2501} + 1 < -1.00000000064135 < -1$$
Thus, $h^{5001} - 2h^{2501} + 1 < 0$.

9. Total Initial Weight:
The total initial weight of the cows is:
$$\sum_{i=1}^{\infty} \frac{1}{h^i} = \frac{1}{h-1} = \frac{10^9}{159} \approx 6289309$$

10. Weight of a Cow at 100,000,000:
Using Bernoulli's inequality and approximations:
$$h^{100,000,000} > 4.79^{10} > 4.79^3 \cdot 4.79^7 > 109.9 \cdot 12000 > 6289309$$
Thus, $h^{100,000,000} > \frac{10^9}{159}$.

Since the weight of a cow at 100,000,000 exceeds the total initial weight, it is impossible for Farmer John to get a cow to position 100,000,000.