Problem 1513

1. We start with the given equation:
$$1 + 2 + \dots + (n - 1) = (n + 1) + (n + 2) + \dots + (n + r)$$

2. The sum of the first $n-1$ positive integers on the left-hand side (LHS) can be expressed using the formula for the sum of an arithmetic series:
$$1 + 2 + \dots + (n - 1) = \frac{(n-1)n}{2}$$

3. The sum of $r$ consecutive integers starting from $n+1$ on the right-hand side (RHS) can be expressed as:
$$(n + 1) + (n + 2) + \dots + (n + r) = \frac{r}{2} \left(2n + r + 1\right)$$

4. Equating the LHS and RHS, we get:
$$\frac{(n-1)n}{2} = \frac{r}{2} \left(2n + r + 1\right)$$

5. Multiplying both sides by 2 to clear the fractions:
$$n(n-1) = r(2n + r + 1)$$

6. Suppose $n$ is a prime number, let $n = p$ where $p$ is prime. Then the equation becomes:
$$p(p-1) = r(2p + r + 1)$$

7. Since $p$ is prime, it must divide one of the factors on the RHS, i.e., $r$ or $2p + r + 1$.

8. Note that $p < 2p + r + 1$, so $p-1 > r$. This implies $p$ cannot divide $r$ because $r < p-1$.

9. Therefore, $p$ must divide $2p + r + 1$. This implies:
$$2p + r + 1 \equiv 0 \pmod{p}$$
Simplifying, we get:
$$r + 1 \equiv 0 \pmod{p} \implies r \equiv -1 \pmod{p}$$

10. Since $r$ is a positive integer, this implies that $r$ must be at least $p-1$.

11. If $r \geq p-1$, substituting $r = p-1$ into the original equation, we get:
$$\text{RHS} - \text{LHS} \geq [(p+1) + (p+2) + \dots + (2p-1)] - [1 + 2 + \dots + (p-1)]$$
Simplifying, we get:
$$= [(p+1) - 1] + [(p+2) - 2] + \dots + [(2p-1) - (p-1)]$$
$$= p + p + \dots + p$$
$$= p(p-1)$$
$$> 0$$
This is a contradiction because the RHS cannot be greater than the LHS.

12. Hence, $n$ cannot be a prime number. Therefore, $n$ must be a composite number.

13. Additionally, for $n = 1$, the LHS is 0 and the RHS is positive, so $LHS < RHS$.