Let $a_1,a_2,\cdots,a_{100}\geq  0$ such that $\max\{a_{i-1}+a_i,a_i+a_{i+1}\}\geq i $ for any $2\leq i\leq 99.$ Find the minimum of $a_1+a_2+\cdots+a_{100}.$
Given the sequence \(a_1, a_2, \ldots, a_{100} \geq 0\) such that \(\max\{a_{i-1} + a_i, a_i + a_{i+1}\} \geq i\) for any \(2 \leq i \leq 99\), we need to find the minimum value of \(a_1 + a_2 + \cdots + a_{100}\).

1. **Analyzing the given condition:**
   \[
   \max\{a_{i-1} + a_i, a_i + a_{i+1}\} \geq i
   \]
   This implies that either \(a_{i-1} + a_i \geq i\) or \(a_i + a_{i+1} \geq i\) (or both).

2. **Constructing a sequence that satisfies the condition:**
   Let's consider a sequence where \(a_i = \frac{i}{2}\) for \(i = 1, 2, \ldots, 100\).

3. **Verifying the sequence:**
   For \(2 \leq i \leq 99\):
   \[
   a_{i-1} + a_i = \frac{i-1}{2} + \frac{i}{2} = \frac{2i-1}{2}
   \]
   \[
   a_i + a_{i+1} = \frac{i}{2} + \frac{i+1}{2} = \frac{2i+1}{2}
   \]
   We need to check if:
   \[
   \max\left\{\frac{2i-1}{2}, \frac{2i+1}{2}\right\} \geq i
   \]
   Clearly, \(\frac{2i+1}{2} \geq i\) for all \(i\), since:
   \[
   \frac{2i+1}{2} = i + \frac{1}{2} \geq i
   \]

4. **Calculating the sum:**
   \[
   a_1 + a_2 + \cdots + a_{100} = \frac{1}{2} + \frac{2}{2} + \cdots + \frac{100}{2} = \frac{1 + 2 + \cdots + 100}{2}
   \]
   Using the formula for the sum of the first \(n\) natural numbers:
   \[
   1 + 2 + \cdots + 100 = \frac{100 \cdot 101}{2} = 5050
   \]
   Therefore:
   \[
   a_1 + a_2 + \cdots + a_{100} = \frac{5050}{2} = 2525
   \]

5. **Checking for a lower sum:**
   We need to check if there is a sequence that gives a lower sum while satisfying the condition. Consider the sequence where \(a_i = 0\) for all \(i\). This sequence does not satisfy the condition since \(\max\{a_{i-1} + a_i, a_i + a_{i+1}\} = 0\) which is not \(\geq i\).

Thus, the minimum sum of \(a_1 + a_2 + \cdots + a_{100}\) that satisfies the given condition is \(2525\).

The final answer is \(\boxed{2525}\).