Problem 1216

1. Understanding the Problem:
We need to find the maximum value of the sum
$$S(n) = |a_1 - a_2| + |a_2 - a_3| + \cdots + |a_{n-1} - a_n|$$
where $a_1, a_2, \ldots, a_n$ is a permutation of the numbers $1, 2, \ldots, n$.

2. Analyzing the Absolute Differences:
To maximize $S(n)$, we need to maximize each term $|a_i - a_{i+1}|$. The maximum difference between any two numbers in the set $\{1, 2, \ldots, n\}$ is $n-1$.

3. Constructing the Permutation:
Consider the permutation that alternates between the largest and smallest remaining numbers. For example, for $n = 4$, the permutation would be $1, 4, 2, 3$. This ensures that the differences are maximized.

4. General Pattern:
For a general $n$, the permutation can be constructed as follows:
- Start with the smallest number, then the largest, then the second smallest, then the second largest, and so on.
- This can be written as $1, n, 2, n-1, 3, n-2, \ldots$.

5. Calculating the Sum:
Let's calculate $S(n)$ for this permutation:
- For $n = 4$, the permutation is $1, 4, 2, 3$:
$$S(4) = |1 - 4| + |4 - 2| + |2 - 3| = 3 + 2 + 1 = 6$$
- For $n = 5$, the permutation is $1, 5, 2, 4, 3$:
$$S(5) = |1 - 5| + |5 - 2| + |2 - 4| + |4 - 3| = 4 + 3 + 2 + 1 = 10$$

6. General Formula:
The sum $S(n)$ for the optimal permutation is:
$$S(n) = (n-1) + (n-2) + \cdots + 1 = \sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}$$

7. Conclusion:
The maximum value of $S(n)$ is given by:
$$S(n) = \frac{(n-1)n}{2}$$

The final answer is $\boxed{\frac{(n-1)n}{2}}$