The number 2021 is fantabulous. For any positive integer $m$, if any element of the set $\{m, 2m+1, 3m\}$ is fantabulous, then all the elements are fantabulous. Does it follow that the number $2021^{2021}$ is fantabulous?
1. **Understanding the Problem:**
   We need to determine if the number \(2021^{2021}\) is fantabulous given the properties of fantabulous numbers. Specifically, if any element of the set \(\{m, 2m+1, 3m\}\) is fantabulous, then all elements of the set are fantabulous.

2. **Initial Claims:**
   - Claim: \(m\) is fantabulous if \(2m+1\) is fantabulous, and \(2m+1\) is fantabulous if \(m\) is fantabulous.
   - Claim: \(m\) is fantabulous if \(2m\) is fantabulous, and \(2m\) is fantabulous if \(m\) is fantabulous.

3. **Proof of Claims:**
   - For the first claim, if \(2m+1\) is fantabulous, then by the given property, \(m\) must also be fantabulous. Similarly, if \(m\) is fantabulous, then \(2m+1\) must be fantabulous.
   - For the second claim, consider the chain \(2m \to 4m+1 \to 12m+3 \to 6m+1 \to 3m \to m\). This chain shows that if \(2m\) is fantabulous, then \(m\) is fantabulous. The reverse chain shows that if \(m\) is fantabulous, then \(2m\) is fantabulous.

4. **Generalization to All Positive Integers:**
   - For any integer \(n > 1\), if \(n\) is even, we can divide by 2 to find another fantabulous number. If \(n\) is odd, we can subtract 1 and then divide by 2 to find another fantabulous number. This process can be repeated until we reach 1.

5. **Base Case and Inductive Step:**
   - Base Case: \(1\) is fantabulous.
   - Inductive Step: Suppose all \(k < n\) are fantabulous. If \(n\) is even, \(\frac{n}{2}\) is fantabulous, so \(n\) is fantabulous. If \(n > 1\) is odd, \(\frac{n-1}{2}\) is fantabulous, so \(n\) is fantabulous.

6. **Conclusion:**
   By induction, every positive integer is fantabulous. Therefore, \(2021^{2021}\) is also fantabulous.

The final answer is \(\boxed{\text{yes}}\)