The problem is asking for which values of $n$, the sum of the integers from 1 to $n-1$, denoted by $1 + 2 + \cdots + (n-1)$, is congruent to 0 modulo $n$, i.e.,

$1 + 2 + \cdots + (n-1) \equiv 0 \pmod{n}.$

This sum is the arithmetic series sum:

$S = 1 + 2 + \cdots + (n-1) = \frac{(n-1)n}{2}.$

We want this sum to be congruent to 0 modulo $n$:

$\frac{(n-1)n}{2} \equiv 0 \pmod{n}.$

This simplifies to the condition:

$\frac{(n-1)n}{2} \equiv 0 \pmod{n}.$

Now, observe that $\frac{(n-1)n}{2} \equiv 0 \pmod{n}$ will hold if and only if $n$ divides $\frac{(n-1)n}{2}$. For this to happen, $n$ must divide $n-1$, which is impossible for any $n \geq 2$, unless $n$ divides the factor $2$ as well.

Thus, $n$ must divide 2. The divisors of 2 are $n = 1$ and $n = 2$.

Conclusion:

The only values of $n$ that satisfy the condition are $n = 1$ and $n = 2$.

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Would you like more details on any part of this explanation, or do you have any further questions?

Here are 5 related questions to expand on this concept:

1. What is the general formula for the sum of an arithmetic series?
2. How does modular arithmetic simplify calculations in number theory?
3. Why can't $n \geq 3$ satisfy the condition in this problem?
4. What are the properties of divisors in number theory?
5. How can we generalize this result to other sums with different moduli?

Tip: Modular arithmetic often simplifies number theory problems by focusing only on the remainder, ignoring the full value of numbers. This can reveal useful patterns or simplifications.