Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ,$ find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}.$
1. **Express \( z \) in exponential form:**
   Given \( z + \frac{1}{z} = 2 \cos 3^\circ \), we can write \( z \) as \( z = e^{i\theta} \). Therefore, \( \frac{1}{z} = e^{-i\theta} \).

2. **Simplify the given equation:**
   \[
   e^{i\theta} + e^{-i\theta} = 2 \cos 3^\circ
   \]
   Using Euler's formula, \( e^{i\theta} + e^{-i\theta} = 2 \cos \theta \). Thus, we have:
   \[
   2 \cos \theta = 2 \cos 3^\circ
   \]
   Dividing both sides by 2, we get:
   \[
   \cos \theta = \cos 3^\circ
   \]

3. **Determine the possible values of \( \theta \):**
   Since \( \cos \theta = \cos 3^\circ \), \( \theta \) can be \( 3^\circ \) or \( -3^\circ \) (or any angle coterminal with these, i.e., \( \theta = 3^\circ + 360^\circ k \) or \( \theta = -3^\circ + 360^\circ k \) for any integer \( k \)).

4. **Calculate \( z^{2000} + \frac{1}{z^{2000}} \):**
   We need to find \( e^{i \cdot 2000 \theta} + e^{-i \cdot 2000 \theta} \). Using \( \theta = 3^\circ \):
   \[
   e^{i \cdot 2000 \cdot 3^\circ} + e^{-i \cdot 2000 \cdot 3^\circ} = e^{i \cdot 6000^\circ} + e^{-i \cdot 6000^\circ}
   \]

5. **Simplify the angle:**
   Note that \( 6000^\circ \) can be reduced modulo \( 360^\circ \):
   \[
   6000 \mod 360 = 240^\circ
   \]
   Therefore:
   \[
   e^{i \cdot 6000^\circ} = e^{i \cdot 240^\circ}
   \]
   and
   \[
   e^{-i \cdot 6000^\circ} = e^{-i \cdot 240^\circ}
   \]

6. **Evaluate the expression:**
   \[
   e^{i \cdot 240^\circ} + e^{-i \cdot 240^\circ} = 2 \cos 240^\circ
   \]
   Since \( \cos 240^\circ = \cos (180^\circ + 60^\circ) = -\cos 60^\circ = -\frac{1}{2} \):
   \[
   2 \cos 240^\circ = 2 \left( -\frac{1}{2} \right) = -1
   \]

7. **Find the least integer greater than the result:**
   The value of \( z^{2000} + \frac{1}{z^{2000}} \) is \( -1 \). The least integer greater than \( -1 \) is \( 0 \).

The final answer is \( \boxed{0} \).