Assume real numbers $a_i,b_i\,(i=0,1,\cdots,2n)$ satisfy the following conditions:
(1) for $i=0,1,\cdots,2n-1$, we have $a_i+a_{i+1}\geq 0$;
(2) for $j=0,1,\cdots,n-1$, we have $a_{2j+1}\leq 0$;
(2) for any integer $p,q$, $0\leq p\leq q\leq n$, we have $\sum_{k=2p}^{2q}b_k>0$.
Prove that $\sum_{i=0}^{2n}(-1)^i a_i b_i\geq 0$, and determine when the equality holds.
1. **Define new sequences**: Let \( c_i = (-1)^i a_i \) for \( i = 0, 1, \ldots, 2n \). This transformation helps in simplifying the given conditions. Also, define \( B_i = \sum_{k=2n-i+1}^{2n} b_k \).

2. **Base case for induction**: Consider \( n = 0 \). In this case, we need to show that \( c_0 b_0 \geq 0 \). Since \( c_0 = a_0 \) and \( b_0 > 0 \) (from the third condition), it follows that \( c_0 b_0 = a_0 b_0 \geq 0 \).

3. **Induction hypothesis**: Assume that for some \( n \), the inequality \( \sum_{i=0}^{2n} c_i b_i \geq 0 \) holds.

4. **Induction step**: We need to show that \( \sum_{i=0}^{2(n+1)} c_i b_i \geq 0 \). Consider the sum:
   \[
   \sum_{i=0}^{2(n+1)} c_i b_i = \sum_{i=0}^{2n+2} c_i b_i
   \]
   We can split this sum into two parts:
   \[
   \sum_{i=0}^{2n+2} c_i b_i = \sum_{i=0}^{2n} c_i b_i + c_{2n+1} b_{2n+1} + c_{2n+2} b_{2n+2}
   \]

5. **Using Abel's transformation**: Abel's transformation is a technique used to transform sums to make them easier to handle. We apply it to the sum \( \sum_{i=0}^{2n+2} c_i b_i \):
   \[
   \sum_{i=0}^{2n+2} c_i b_i = \sum_{i=0}^{2n+1} (c_i - c_{i+1}) B_{i+1} + c_{2n+2} B_{2n+2}
   \]
   where \( B_i = \sum_{k=2n-i+1}^{2n} b_k \).

6. **Analyzing the terms**: From the given conditions:
   - \( a_i + a_{i+1} \geq 0 \) implies \( c_i - c_{i+1} \geq 0 \) for \( i = 0, 1, \ldots, 2n-1 \).
   - \( a_{2j+1} \leq 0 \) implies \( c_{2j+1} \leq 0 \) for \( j = 0, 1, \ldots, n-1 \).
   - \( \sum_{k=2p}^{2q} b_k > 0 \) for \( 0 \leq p \leq q \leq n \) implies \( B_i > 0 \) for \( i = 1, 2, \ldots, 2n+2 \).

7. **Combining the results**: Since \( c_i - c_{i+1} \geq 0 \) and \( B_i > 0 \), each term in the sum \( \sum_{i=0}^{2n+1} (c_i - c_{i+1}) B_{i+1} \) is non-negative. Additionally, \( c_{2n+2} B_{2n+2} \geq 0 \) because \( c_{2n+2} = (-1)^{2n+2} a_{2n+2} = a_{2n+2} \) and \( B_{2n+2} > 0 \).

8. **Conclusion**: Therefore, the sum \( \sum_{i=0}^{2(n+1)} c_i b_i \geq 0 \) holds, completing the induction step.

9. **Equality condition**: The equality holds if and only if \( a_0 = a_1 = \cdots = a_{2n} = 0 \), which implies \( c_i = 0 \) for all \( i \).

\(\blacksquare\)

The final answer is \( \boxed{ \sum_{i=0}^{2n} (-1)^i a_i b_i \geq 0 } \)