11. 4. Given a positive integer $n(n \geqslant 3)$. Find the smallest positive integer $k$ such that the following conclusion always holds.

For any $n$ points $A_{i}=\left(x_{i}, y_{i}\right)$ in the plane, no three of which are collinear, and any $n$ real numbers $c_{i}(1 \leqslant i \leqslant n)$, there exists a bivariate real polynomial $P(x, y)$ of degree not exceeding $k$, satisfying
$$
P\left(x_{i}, y_{i}\right)=c_{i}(i=1,2, \cdots, n) .
$$
11.4. $k=\left[\frac{n}{2}\right]$.

First, prove a lemma.
Lemma For any $n$ given points $A_{i}=\left(x_{i}, y_{i}\right)(1 \leqslant i \leqslant n)$ in the plane, no three of which are collinear, there exists a polynomial $P(x, y)$ of degree not greater than $\left[\frac{n}{2}\right]$, satisfying
$$
P\left(x_{n}, y_{n}\right)=1, P\left(x_{i}, y_{i}\right)=0(1 \leqslant i \leqslant n-1) .
$$

Proof There exist $d=\left[\frac{n}{2}\right]$ lines such that $A_{n}$ is not on these lines, while $A_{1}, A_{2}, \cdots, A_{n-1}$ belong to at least one of them (in fact, when $n$ is even, $A_{1} A_{2}$, $A_{3} A_{4}, \cdots, A_{n-3} A_{n-2}, A_{1} A_{n-1}$; when $n$ is odd, $A_{1} A_{2}, A_{3} A_{4}, \cdots, A_{n-2} A_{n-1}$).

Let $k_{i} x+l_{i} y+m_{i}=0(1 \leqslant i \leqslant d)$ be the equations of these lines. Then the polynomial
$$
Q(x, y)=\prod_{i=1}^{d} \frac{\left(k_{i} x+l_{i} y+m_{i}\right)}{\left(k_{i} x_{n}+l_{i} y_{n}+m_{i}\right)}
$$

satisfies the conditions.
Return to the original problem.
First, prove: $k=\left[\frac{n}{2}\right]$ satisfies the requirement.
For $i=1,2, \cdots, n$, by the lemma, there exists a $k$-degree polynomial $P_{i}(x, y)$, such that $P_{i}$ is 0 at all points $A_{1}, A_{2}, \cdots, A_{n}$ except $A_{i}$, and $P_{i}\left(x_{i}, y_{i}\right)=1$. Then
$$
P(x, y)=\sum_{i=1}^{n} c_{i} P_{i}(x, y)
$$

satisfies the requirement.
Next, prove: $k<\left[\frac{n}{2}\right]$ the conclusion does not hold.
Consider the points $A_{i}\left(i, i^{2}\right), 1 \leqslant i \leqslant n, c_{1}=c_{2}=\cdots=$ $c_{n-1}=0, c_{n}=1$. Then $A_{1}, A_{2}, \cdots, A_{n}$ are such that no three are collinear.

Assume there exists a polynomial $P(x, y)$ of degree $k$, satisfying $P\left(x_{i}, y_{i}\right)=c_{i}$. Let $Q(x)=P\left(x, x^{2}\right)$.
Then the degree of $Q$ does not exceed $2 k$, and
$$
Q(i)=0(i=1,2, \cdots, n-1), Q(n)=1 \text {. }
$$

Hence $2 k \geqslant n-1$, which implies $k \geqslant\left[\frac{n}{2}\right]$.