Three, (20 points) Place a real number in each cell of a $4 \times 4$ grid paper, 
such that the sum of the four numbers in each row, each column, and each diagonal equals a constant $k$. Find the sum of the numbers in the four corners of this $4 \times 4$ grid paper.
Three, we use $a_{ij}$ to denote the number in the $i$-th row and $j$-th column of a $4 \times 4$ grid $(1 \leqslant i, j \leqslant 4)$. From the given information, we have
$$
\left\{\begin{array}{l}
\sum_{i=1}^{4} a_{i 2}+\sum_{i=1}^{4} a_{i 3}=2 k, \\
\sum_{j=1}^{4} a_{1 j}+\sum_{j=1}^{4} a_{4 j}=2 k .
\end{array}\right.
$$

Subtracting the two equations and rearranging, we get
$$
a_{11}+a_{14}+a_{41}+a_{44}=a_{22}+a_{23}+a_{32}+a_{33} \text {. }
$$

Given that the sum of the four numbers on the diagonal is $k$, we have
$$
\begin{array}{l}
\left(a_{11}+a_{22}+a_{33}+a_{44}\right)+\left(a_{41}+a_{32}+\dot{a}_{23}\right. \\
\left.+a_{14}\right)=2 k .
\end{array}
$$

Therefore, $a_{11}+a_{14}+a_{41}+a_{44}=k$,
which means the sum of the numbers in the four corners is $k$.