Let $p$ is a prime and $p\equiv 2\pmod 3$. For $\forall a\in\mathbb Z$, if
$$p\mid \prod\limits_{i=1}^p(i^3-ai-1),$$then $a$ is called a "GuGu" number. How many "GuGu" numbers are there in the set $\{1,2,\cdots ,p\}?$

(We are allowed to discuss now. It is after 00:00 Feb 14 Beijing Time)
1. **Initial Assumptions and Simplifications:**
   - Given \( p \) is a prime such that \( p \equiv 2 \pmod{3} \).
   - We need to find the number of "GuGu" numbers in the set \(\{1, 2, \ldots, p\}\).
   - A number \( a \) is called a "GuGu" number if \( p \mid \prod_{i=1}^p (i^3 - ai - 1) \).

2. **Considering the Polynomial:**
   - The polynomial \( f(x) = x^3 - ax - 1 \) has \( p \) roots modulo \( p \).
   - By Fermat's Little Theorem, for any integer \( i \), \( i^p \equiv i \pmod{p} \).

3. **Roots and Polynomial Properties:**
   - We need to find the values of \( a \) such that \( f(x) \) has a root modulo \( p \).
   - This means we need to find \( a \) such that there exists \( x \) with \( x^3 - ax - 1 \equiv 0 \pmod{p} \).

4. **Counting Solutions:**
   - Let \( S_A = \{ x \mid x^3 - Ax - 1 \equiv 0 \pmod{p} \} \).
   - We need to count the number of distinct values \( A \) such that \( |S_A| \neq 0 \).

5. **Using the Given Conditions:**
   - We know \( p \equiv 2 \pmod{3} \), which implies \( p = 3k + 2 \) for some integer \( k \).
   - We need to count the number of distinct \( A \) modulo \( p \) such that \( x^3 - Ax - 1 \equiv 0 \) has solutions.

6. **Analyzing the Polynomial:**
   - For each \( x \), \( x^3 - Ax - 1 \equiv 0 \) can be rewritten as \( A \equiv \frac{x^3 - 1}{x} \pmod{p} \).
   - We need to count the number of distinct values of \( \frac{x^3 - 1}{x} \) modulo \( p \).

7. **Counting Distinct Values:**
   - Since \( p \equiv 2 \pmod{3} \), the polynomial \( x^3 - 1 \) has exactly \( 3 \) distinct roots modulo \( p \).
   - Therefore, the number of distinct values of \( \frac{x^3 - 1}{x} \) modulo \( p \) is \( \frac{2p - 1}{3} \).

8. **Conclusion:**
   - The number of "GuGu" numbers in the set \(\{1, 2, \ldots, p\}\) is \( \frac{2p - 1}{3} \).

The final answer is \(\boxed{\frac{2p - 1}{3}}\)