1. **Problem statement:** Find the sum of all positive integers $n \leq 200$ such that the sum of all distinct positive divisors of $n$ equals $2n - 1$. 2. **Understanding the problem:** Let $\sigma(n)$ denote the sum of all positive divisors of $n$. The problem states: $$\sigma(n) = 2n - 1$$ 3. **Key insight:** For any positive integer $n$, the sum of divisors $\sigma(n)$ is at least $n + 1$ (since $1$ and $n$ are divisors). The equation can be rearranged: $$\sigma(n) = 2n - 1 \implies \sigma(n) - n = n - 1$$ 4. The term $\sigma(n) - n$ is the sum of the *proper divisors* of $n$ (all divisors excluding $n$ itself). So: $$\text{sum of proper divisors of } n = n - 1$$ 5. **Interpretation:** The sum of proper divisors of $n$ is exactly one less than $n$. This is a special property. 6. **Check for prime powers:** For $n = p^k$ where $p$ is prime, the sum of divisors is: $$\sigma(p^k) = 1 + p + p^2 + \cdots + p^k = \frac{p^{k+1} - 1}{p - 1}$$ 7. The sum of proper divisors is: $$\sigma(p^k) - p^k = \frac{p^{k+1} - 1}{p - 1} - p^k$$ 8. Set this equal to $n - 1 = p^k - 1$: $$\frac{p^{k+1} - 1}{p - 1} - p^k = p^k - 1$$ 9. Rearranged: $$\frac{p^{k+1} - 1}{p - 1} = 2p^k - 1$$ 10. Multiply both sides by $p - 1$: $$p^{k+1} - 1 = (2p^k - 1)(p - 1)$$ 11. Expand right side: $$p^{k+1} - 1 = 2p^{k+1} - 2p^k - p + 1$$ 12. Bring all terms to one side: $$p^{k+1} - 1 - 2p^{k+1} + 2p^k + p - 1 = 0$$ 13. Simplify: $$-p^{k+1} + 2p^k + p - 2 = 0$$ 14. Multiply both sides by $-1$: $$p^{k+1} - 2p^k - p + 2 = 0$$ 15. Factor: $$p^{k}(p - 2) - (p - 2) = (p - 2)(p^k - 1) = 0$$ 16. So either $p = 2$ or $p^k = 1$ (impossible for $k \geq 1$). Thus, $p=2$. 17. Therefore, $n$ must be a power of 2: $n = 2^k$. 18. Check if all powers of 2 satisfy the original condition: Sum of divisors of $2^k$ is: $$\sigma(2^k) = 1 + 2 + 4 + \cdots + 2^k = 2^{k+1} - 1$$ 19. Check if $\sigma(2^k) = 2n - 1$: $$2n - 1 = 2 \cdot 2^k - 1 = 2^{k+1} - 1$$ 20. This matches exactly. 21. So all powers of 2 satisfy the condition. 22. Find all powers of 2 less than or equal to 200: $$2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32, 2^6 = 64, 2^7 = 128, 2^8 = 256 > 200$$ 23. So valid $n$ are: $$1, 2, 4, 8, 16, 32, 64, 128$$ 24. Sum these: $$1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255$$ 25. **Final answer:** 255 **Answer choice:** (E) 255