$36 \cdot 2$ a $3 \times 3$ array of numbers
$$
\left(\begin{array}{lll}
4 & 9 & 2 \\
8 & 1 & 6 \\
3 & 5 & 7
\end{array}\right)
$$

Each row and each column has the same sum. To make the sums of each row and each column different, the minimum number of numbers that need to be changed is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(E) 5 .
(48th American High School Mathematics Examination, 1997)
[Solution] Since each number is in one row and one column, changing one number can alter the sum of the numbers in the row and column where the number is located, but they remain equal to each other.

Changing two numbers can affect at most 4 row or column sums (2 rows, 2 columns), so to change 6 row or column sums, at least three numbers need to be changed.

These three numbers can be in the same row (or the same column), in two different columns of the same row (or two different rows of the same column), or in different rows and different columns:
$$
\left(\begin{array}{lll}
* & * & * \\
& & \\
& & * \\
* & \\
&
\end{array}\right) \text { or }\left(\begin{array}{lll}
* & & \\
& & \\
& & \\
& & \\
& & *
\end{array}\right)
$$
(Note that swapping two rows or two columns in the number array does not affect the equality of the sums)
Clearly, the above changes still cannot make the six row or column sums of the number array all unequal. Changing four numbers is feasible, for example: $( \left.\begin{array}{ll}* & * \\ & * \\ & \end{array}\right)$ Therefore, the answer is $(D)$.