Find the least number $ c$ satisfyng the condition $\sum_{i=1}^n {x_i}^2\leq cn$  
and all real numbers $x_1,x_2,...,x_n$ are greater than or equal to $-1$ such that $\sum_{i=1}^n {x_i}^3=0$
To find the least number \( c \) satisfying the condition \(\sum_{i=1}^n {x_i}^2 \leq cn\) for all real numbers \(x_1, x_2, \ldots, x_n \geq -1\) such that \(\sum_{i=1}^n {x_i}^3 = 0\), we can proceed as follows:

1. **Define the Functions and Constraints:**
   - Define \( g: \mathbb{R}^n \rightarrow \mathbb{R} \) by \( g(x) = \sum_{i=1}^n x_i^3 \).
   - Define \( f: \mathbb{R}^n \rightarrow \mathbb{R} \) by \( f(x) = \sum_{i=1}^n x_i^2 \).
   - Consider the set \( A = \{ x \in \mathbb{R}^n \mid x_i \geq -1 \text{ and } g(x) = 0 \} \).

2. **Compactness and Continuity:**
   - The set \( A \) is closed and bounded, hence compact.
   - The functions \( f \) and \( g \) are continuous.

3. **Lagrange Multipliers:**
   - Use the method of Lagrange multipliers to find the extrema of \( f \) subject to the constraint \( g(x) = 0 \).
   - The Lagrangian is \( \mathcal{L}(x, \lambda) = \sum_{i=1}^n x_i^2 + \lambda \left( \sum_{i=1}^n x_i^3 \right) \).

4. **Solve the System of Equations:**
   - The partial derivatives are:
     \[
     \frac{\partial \mathcal{L}}{\partial x_i} = 2x_i + 3\lambda x_i^2 = 0 \quad \text{for all } i.
     \]
   - This gives \( x_i (2 + 3\lambda x_i) = 0 \). Hence, \( x_i = 0 \) or \( \lambda = -\frac{2}{3x_i} \).

5. **Case Analysis:**
   - If \( x_i = 0 \) for all \( i \), then \( \sum_{i=1}^n x_i^3 = 0 \) is trivially satisfied, but this does not help in maximizing \( f \).
   - Assume \( x_i \neq 0 \). Then \( x_i = \pm k \) for some constant \( k \).

6. **Symmetry and Simplification:**
   - If \( n \) is even, we can have \( n/2 \) values of \( x_i = k \) and \( n/2 \) values of \( x_i = -k \).
   - This satisfies \( \sum_{i=1}^n x_i^3 = 0 \) because the positive and negative terms cancel out.

7. **Maximize \( f \):**
   - For \( x_i = k \) and \( x_i = -k \), we have:
     \[
     \sum_{i=1}^n x_i^2 = \frac{n}{2} k^2 + \frac{n}{2} k^2 = nk^2.
     \]
   - The constraint \( x_i \geq -1 \) implies \( k \leq 1 \).

8. **Boundary Case:**
   - Consider the case where \( x_i = 1 \) for \( m \) values and \( x_i = -1 \) for \( n-m \) values.
   - The condition \( \sum_{i=1}^n x_i^3 = 0 \) gives:
     \[
     m - (n-m) = 0 \implies m = \frac{n}{2}.
     \]
   - Then:
     \[
     \sum_{i=1}^n x_i^2 = m \cdot 1^2 + (n-m) \cdot (-1)^2 = \frac{n}{2} + \frac{n}{2} = n.
     \]

9. **Conclusion:**
   - The maximum value of \( \sum_{i=1}^n x_i^2 \) is \( n \).
   - Therefore, the least number \( c \) such that \( \sum_{i=1}^n x_i^2 \leq cn \) is \( c = 1 \).

The final answer is \( \boxed{1} \).