Find largest possible constant $M$ such that, for any sequence $a_n$, $n=0,1,2,...$ of real numbers, that satisfies the conditions :
i) $a_0=1$, $a_1=3$
ii) $a_0+a_1+...+a_{n-1} \ge 3 a_n - a_{n+1}$ for any integer $n\ge 1$
to be true that
$$\frac{a_{n+1}}{a_n} >M$$ for any integer $n\ge 0$.
To find the largest possible constant \( M \) such that for any sequence \( \{a_n\} \) of real numbers satisfying the given conditions, the inequality \( \frac{a_{n+1}}{a_n} > M \) holds for any integer \( n \geq 0 \), we proceed as follows:

1. **Initial Conditions and Sequence Definition:**
   - Given \( a_0 = 1 \) and \( a_1 = 3 \).
   - The sequence satisfies \( a_0 + a_1 + \cdots + a_{n-1} \geq 3a_n - a_{n+1} \) for any integer \( n \geq 1 \).

2. **Summation Notation:**
   - Define \( S_n = a_0 + a_1 + \cdots + a_n \).
   - The given condition can be rewritten as \( S_{n-1} \geq 3a_n - a_{n+1} \).

3. **Inductive Step:**
   - We need to show that \( S_n > 2S_{n-1} \) for all \( n \geq 1 \).
   - Base case: \( S_0 = 1 \) and \( S_1 = 4 \) (since \( S_1 = a_0 + a_1 = 1 + 3 = 4 \)).
   - Assume \( S_k > 2S_{k-1} \) for some \( k \geq 1 \).

4. **Using the Given Condition:**
   - From \( S_{n-1} \geq 3a_n - a_{n+1} \), we have:
     \[
     S_{n-1} \geq 3(S_n - S_{n-1}) - (S_{n+1} - S_n)
     \]
   - Simplifying, we get:
     \[
     S_{n-1} \geq 3S_n - 3S_{n-1} - S_{n+1} + S_n
     \]
     \[
     S_{n+1} \geq 4S_n - 4S_{n-1}
     \]

5. **Inductive Conclusion:**
   - From the inductive hypothesis \( S_k > 2S_{k-1} \), we have:
     \[
     S_{k+1} \geq 4S_k - 4S_{k-1} > 2S_k
     \]
   - Therefore, by induction, \( S_n > 2S_{n-1} \) for all \( n \geq 1 \).

6. **Ratio of Consecutive Terms:**
   - From \( S_n > 2S_{n-1} \), we have:
     \[
     a_{n+1} - 2a_n \geq a_n - S_{n-1} = S_n - 2S_{n-1} > 0
     \]
   - Thus, \( \frac{a_{n+1}}{a_n} > 2 \).

7. **Upper Bound for \( M \):**
   - To show that \( M > 2 \) does not hold, consider \( a_{n+1} = 3a_n - S_{n-1} \).
   - Then \( S_{n+1} = 4S_n - 4S_{n-1} \).
   - Given \( S_0 = 1 \) and \( S_1 = 4 \), we have \( S_n = 2^n + n \cdot 2^n = (n+2) \cdot 2^n \).
   - Therefore, \( a_n = S_n - S_{n-1} = (n+2) \cdot 2^{n-1} \).

8. **Limit Calculation:**
   - Calculate the limit:
     \[
     \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{(n+3) \cdot 2^n}{(n+2) \cdot 2^{n-1}} = \lim_{n \to \infty} 2 \cdot \frac{n+3}{n+2} = 2
     \]
   - Hence, \( M = 2 \).

The final answer is \( \boxed{ M = 2 } \).