$$
\text { Three. (50 points) Given }(2 k+1) \times(2 k+1)
$$
$\left(k \in \mathbf{N}_{*}\right)$, each element in the square grid is a real number with an absolute value no greater than 1, and the sum of all elements in the grid is equal to 0. Find the smallest non-negative real number $C$, such that in every such grid, there must be a row or a column whose elements' sum has an absolute value no greater than $C$.
Three, first, the specialized square grid table
$$
a_{i j}=\left\{\begin{array}{ll}
1, & i, j \leqslant k+1 ; \\
-1, & i, j>k+1 ; \\
-\frac{2 k+1}{2 k(k+1)}, & \text { other cases. }
\end{array}\right.
$$

In this grid, the sum of the elements in the first $k+1$ rows and the first $k+1$ columns is all equal to
$$
k+1-\frac{2 k+1}{2(k+1)}=k+\frac{1}{2 k+2},
$$

and the absolute value of the sum of the elements in the last $k$ rows and the last $k$ columns is all equal to $k+\frac{2 k+1}{2 k}$.
Therefore, $C \geqslant k+\frac{1}{2 k+2}$.
Firstly, assume a grid table satisfies the given conditions, and the absolute value of the sum of each row and each column is greater than $k+\frac{1}{2 k+2}$.

Without loss of generality, assume the sum of the elements in the first $k+1$ rows is all greater than $k+\frac{1}{2 k+2}$.
Thus, the total sum of the first $k+1$ rows is greater than
$$
(k+1)\left(k+\frac{1}{2 k+2}\right).
$$

However, the sum of the elements in each column of the $(k+1) \times n$ sub-table does not exceed $k+1$, so there must be $k+1$ columns with a positive sum of elements, and the sum of the elements in these columns in the original $(2 k+1) \times(2 k+1)$ square grid table is also positive.

Without loss of generality, assume the sum of the elements in the first $k+1$ columns is positive (thus greater than $k+\frac{1}{2 k+2}$).

Notice that the sum of the elements in the upper-left $(k+1) \times(k+1)$ block and the lower-right $k \times k$ block does not exceed $(k+1)^{2}$ and $k^{2}$, respectively. Therefore, the sum of the elements in the grid table is greater than
$$
2(k+1)\left(k+\frac{1}{2 k+2}\right)-(k+1)^{2}-k^{2}=0,
$$

which is a contradiction.
In conclusion, the smallest non-negative real number
$$
C=k+\frac{1}{2 k+2}.
$$