Answer: 30604
We examine the 6 planes, their intersections and the lines between 2 points in one of the three pairs of parallel planes. The expression is equivalent to summing differences in values along all these lines. We examine the planes intersections. There is one cube, $3 \cdot 98$ squares and $3 \cdot 98 \cdot 98$ lines. The minimum value of the difference along a line is 1 . For a square, to minimize the differences we take four consecutive numbers, and the minimum value is 6 . To find the minimum value along a cube, we take 8 consecutive numbers. Since we are taking differences, we can add or subtract any constant to the numbers, so we assume the numbers are 1-8. Examining the cube, we see there's 1 spot where the number is multiplied by $-3,3$ spots where the number is multiplied by $-1,1$ spot where
the number is multiplied by 3 , and 3 spots where the number is multiplied by 1.1 and 8 must go in the corners, and $2,3,4,5$ must go in spots multiplied by $-1,-1,1,1$, respectively. To minimize the differences we put 5 in the final spot multiplied by -1 , and 4 in the spot multiplied by 1 opposite 5. Then the sum of all the differences is 28 , so the minimum for a cube is 28 . So the answer is $28+18(100-2)+3(100-2)^{2}=30000+600+12-36+28=30604$. It is clear that this value can be obtained.
23. $[\mathbf{1 4}]$ Let $S$ be the set of points $(x, y, z)$ in $\mathbb{R}^{3}$ such that $x, y$, and $z$ are positive integers less than or equal to 100 . Let $f$ be a bijective map between $S$ and the $\{1,2, \ldots, 1000000\}$ that satisfies the following property: if $x_{1} \leq x_{2}, y_{1} \leq y_{2}$, and $z_{1} \leq z_{2}$, then $f\left(x_{1}, y_{1}, z_{1}\right) \leq f\left(x_{2}, y_{2}, z_{2}\right)$. Define
$$
\begin{aligned}
A_{i} & =\sum_{j=1}^{100} \sum_{k=1}^{100} f(i, j, k), \\
B_{i} & =\sum_{j=1}^{100} \sum_{k=1}^{100} f(j, i, k), \\
\text { and } C_{i} & =\sum_{j=1}^{100} \sum_{k=1}^{100} f(j, k, i) .
\end{aligned}
$$

Determine the minimum value of $A_{i+1}-A_{i}+B_{j+1}-B_{j}+C_{k+1}-C_{k}$.