Given is a natural number $n \geq 3$. What is the smallest possible value of $k$ if the following statements are true?
For every $n$ points $ A_i = (x_i, y_i) $ on a plane, where no three points are collinear, and for any real numbers $ c_i$  ($1 \le i \le n$) there exists such polynomial $P(x, y)$, the degree of which is no more than $k$, where $ P(x_i, y_i) = c_i $ for every $i = 1, \dots, n$.

(The degree of a nonzero monomial $ a_{i,j} x^{i}y^{j} $ is $i+j$, while the degree of polynomial $P(x, y)$ is the greatest degree of the degrees of its monomials.)
1. **Define the problem and notation:**
   We are given \( n \geq 3 \) points \( A_i = (x_i, y_i) \) on a plane, where no three points are collinear. For any real numbers \( c_i \) (\(1 \le i \le n\)), we need to find the smallest possible value of \( k \) such that there exists a polynomial \( P(x, y) \) of degree at most \( k \) satisfying \( P(x_i, y_i) = c_i \) for every \( i = 1, \dots, n \).

2. **Define the minimum degree function:**
   Let \( f(n) \) denote the minimum degree \( k \) for \( n \) points. We aim to show that \( f(n) = \left\lfloor \frac{n}{2} \right\rfloor \).

3. **Upper bound construction:**
   - We need to show \( f(2n+1) \leq n \). This implies \( f(2n) \leq n \) as well.
   - Consider \( 2n+1 \) distinct points \( A_0, A_1, \dots, A_{2n} \).
   - Pair up points \((A_{2i-1}, A_{2i})\) for \( 1 \leq i \leq n \) and consider the line \(\ell_i\) through them.
   - Construct a polynomial \( P(x, y) \) by multiplying the equations of these lines: \( P(x, y) = \prod_{i=1}^{n} \ell_i(x, y) \).
   - This polynomial \( P(x, y) \) has degree \( \leq n \) and passes through \( A_1, A_2, \dots, A_{2n} \) but not \( A_0 \) since no three points are collinear.

4. **Combining polynomials:**
   - For each point \( A_i \) (\( 0 \leq i \leq 2n \)), construct a polynomial \( P_i(x, y) \) of degree \( \leq n \) that vanishes at all points except \( A_i \).
   - Scale these polynomials appropriately and add them together to obtain the desired polynomial \( P(x, y) \) of degree \( \leq n \) that satisfies \( P(x_i, y_i) = c_i \) for all \( i \).

5. **Lower bound proof:**
   - We need to show \( f(2n) \geq n \). This implies \( f(2n+1) \geq n \) as well.
   - Consider \( 2n \) points lying on a circle, and color one point red.
   - The polynomial \( P(x, y) \) must pass through all points except the red one.
   - By Bezout's theorem, the intersection of the polynomial \( P(x, y) \) and the circle is at most \( 2 \deg P \) points.
   - Since they must intersect in at least \( 2n-1 \) points, it follows that \( \deg P \geq n \).

6. **Conclusion:**
   - Combining the upper and lower bounds, we conclude that \( f(n) = \left\lfloor \frac{n}{2} \right\rfloor \).

\(\blacksquare\)

The final answer is \( \boxed{ \left\lfloor \frac{n}{2} \right\rfloor } \)