Jeffrey stands on a straight horizontal bridge that measures $20000$ meters across. He wishes to place a pole vertically at the center of the bridge so that the sum of the distances from the top of the pole to the two ends of the bridge is $20001$ meters. To the nearest meter, how long of a pole does Jeffrey need?

1. Let the length of the pole be \( h \) meters. The bridge is 20000 meters long, so the center of the bridge is at 10000 meters from either end.
2. By the Pythagorean Theorem, the distance from the top of the pole to each end of the bridge can be expressed as:
   \[
   \sqrt{10000^2 + h^2}
   \]
3. The problem states that the sum of the distances from the top of the pole to the two ends of the bridge is 20001 meters. Therefore, we have:
   \[
   2 \sqrt{10000^2 + h^2} = 20001
   \]
4. Solving for \( h \), we first divide both sides by 2:
   \[
   \sqrt{10000^2 + h^2} = 10000.5
   \]
5. Squaring both sides to eliminate the square root gives:
   \[
   10000^2 + h^2 = 10000.5^2
   \]
6. Calculating \( 10000.5^2 \):
   \[
   10000.5^2 = (10000 + 0.5)^2 = 10000^2 + 2 \cdot 10000 \cdot 0.5 + 0.5^2 = 100000000 + 10000 + 0.25 = 100010000.25
   \]
7. Substituting back into the equation:
   \[
   100000000 + h^2 = 100010000.25
   \]
8. Solving for \( h^2 \):
   \[
   h^2 = 100010000.25 - 100000000 = 10000.25
   \]
9. Taking the square root of both sides:
   \[
   h = \sqrt{10000.25} \approx 100.00125
   \]
10. Rounding to the nearest meter, we get:
   \[
   h \approx 100
   \]

The final answer is \( \boxed{100} \) meters.