A positive integer is [b]superb[/b] if it is the least common multiple of $1,2,\ldots, n$ for some positive integer $n$.
Find all superb $x,y,z$ such that $x+y=z$.
[i]
Proposed by usjl[/i]
1. **Define the superb numbers:**
   Let \( x = \mathrm{lcm}(1, 2, \ldots, a) \), \( y = \mathrm{lcm}(1, 2, \ldots, b) \), and \( z = \mathrm{lcm}(1, 2, \ldots, c) \) for minimal positive integers \( a, b, c \). Without loss of generality, assume \( a < b < c \).

2. **Analyze the powers of 2:**
   The function \(\nu_2(n)\) denotes the highest power of 2 dividing \( n \). For \( x \), \( y \), and \( z \), we have:
   \[
   \nu_2(x) = \nu_2(\mathrm{lcm}(1, 2, \ldots, a)) = \lfloor \log_2 a \rfloor
   \]
   \[
   \nu_2(y) = \nu_2(\mathrm{lcm}(1, 2, \ldots, b)) = \lfloor \log_2 b \rfloor
   \]
   \[
   \nu_2(z) = \nu_2(\mathrm{lcm}(1, 2, \ldots, c)) = \lfloor \log_2 c \rfloor
   \]

3. **Consider the relationship between \( x \), \( y \), and \( z \):**
   Since \( x + y = z \), we need to ensure the powers of 2 align correctly. If there is a power of 2 between \( a \) and \( b \), then \(\nu_2(x) < \nu_2(y)\). Given \( x + y = z \), this would imply \(\nu_2(x) = \nu_2(z)\), which is a contradiction. Therefore, there is no power of 2 between \( a \) and \( b \).

4. **Implications for \( \nu_2(z) \):**
   Since there is no power of 2 between \( a \) and \( b \), \(\nu_2(z)\) must be at least \(\nu_2(x) + 1\). This implies there is a power of 2 between \( b \) and \( c \).

5. **Relationship between \( x \), \( y \), and \( z \):**
   Given \( z \ge 2y \), we conclude \( x = y \) and \( z = 2y \). If \( 2^n \) is the smallest power of 2 at most \( c \), then:
   \[
   2y \mid \mathrm{lcm}(1, 2, \ldots, 2^n) \mid z = 2y
   \]
   This implies \( z = \mathrm{lcm}(1, 2, \ldots, 2^n) \), so \( c = 2^n \).

6. **Determine \( x \) and \( y \):**
   Since \( x = \mathrm{lcm}(1, 2, \ldots, a) \) and \( y = \mathrm{lcm}(1, 2, \ldots, b) \), and \( x = y \), we have:
   \[
   x = \mathrm{lcm}(1, 2, \ldots, 2^n - 1)
   \]

7. **Conclusion:**
   The superb numbers \( x, y, z \) such that \( x + y = z \) are:
   \[
   (x, y, z) = (\mathrm{lcm}(1, 2, \ldots, 2^n - 1), \mathrm{lcm}(1, 2, \ldots, 2^n - 1), \mathrm{lcm}(1, 2, \ldots, 2^n))
   \]
   for any positive integer \( n \).

The final answer is \( \boxed{ (\mathrm{lcm}(1, 2, \ldots, 2^n - 1), \mathrm{lcm}(1, 2, \ldots, 2^n - 1), \mathrm{lcm}(1, 2, \ldots, 2^n)) } \).