Let $a_1, a_2, \cdots , a_n$ be real numbers, with $n\geq  3,$ such that $a_1 + a_2 +\cdots +a_n = 0$ and  $$ 2a_k\leq a_{k-1} + a_{k+1} \ \  \ \text{for} \ \  \  k = 2, 3, \cdots , n-1.$$ Find the least number $\lambda(n),$ such that for all $k\in \{ 1, 2, \cdots, n\} $ it is satisfied that $|a_k|\leq \lambda (n)\cdot \max \{|a_1|, |a_n|\} .$ 
1. **Establishing the Lower Bound:**
   - Consider the sequence where $a_1 = a_n = 1$ and $a_2 = -\frac{n+1}{n-1}$, with $a_2, a_3, \cdots, a_{n-1}$ forming an arithmetic sequence.
   - The sum of the sequence is:
     \[
     a_1 + (n-1) \cdot \frac{a_2 + a_n}{2} = 1 + (n-1) \cdot \frac{-1}{n-1} = 1 - 1 = 0
     \]
   - Here, $|a_2| = \frac{n+1}{n-1}$ and $\max\{|a_1|, |a_n|\} = 1$. This shows that $\lambda(n) \geq \frac{n+1}{n-1}$.

2. **Proving the Upper Bound:**
   - We need to show that $\lambda(n) = \frac{n+1}{n-1}$ satisfies the condition for all $k$.
   - We consider two cases based on the signs of $a_1$ and $a_n$.

3. **Case 1: $a_1a_n \geq 0$**
   - Assume $a_1, a_n \geq 0$ (the case where both are negative is similar).
   - Let $x = \max\{|a_1|, |a_n|\}$. We need to show $|a_k| \leq \frac{n+1}{n-1} \cdot x$ for all $k$.
   - Consider the partial sums:
     \[
     a_1 + a_2 + \cdots + a_k \leq k \cdot \frac{a_1 + a_k}{2} = k \cdot \frac{x + a_k}{2}
     \]
     \[
     a_k + a_{k+1} + \cdots + a_n \leq (n-k+1) \cdot \frac{x + a_k}{2}
     \]
   - Adding these inequalities and using the fact that $a_1 + a_2 + \cdots + a_n = 0$, we get:
     \[
     0 = (a_1 + a_2 + \cdots + a_k) + (a_k + a_{k+1} + \cdots + a_n) - a_k \leq (n+1) \cdot \frac{x + a_k}{2} - a_k
     \]
     \[
     \Rightarrow \frac{n-1}{2} \cdot a_k \geq -\frac{n+1}{2} \cdot x
     \]
     \[
     \Rightarrow a_k \geq -\frac{n+1}{n-1} \cdot x
     \]
   - Similarly, we can show $a_k \leq \frac{n+1}{n-1} \cdot x$.

4. **Case 2: $a_1a_n < 0$**
   - Without loss of generality, assume $a_1 > 0$ and $a_n < 0$.
   - We need to show that $|a_k| \leq \frac{n+1}{n-1} \cdot \max\{|a_1|, |a_n|\}$.
   - Since $a_1 > 0$ and $a_n < 0$, $a_k$ must lie between $a_1$ and $a_n$ due to the concavity condition $2a_k \leq a_{k-1} + a_{k+1}$.
   - This implies $a_k$ cannot exceed $\frac{n+1}{n-1} \cdot \max\{|a_1|, |a_n|\}$.

Thus, we have shown that $\lambda(n) = \frac{n+1}{n-1}$ is both a lower and upper bound, satisfying the given conditions.

The final answer is $\boxed{\frac{n+1}{n-1}}$