Let $f(n):=\mathrm{lcm}(1, 2, \dots, n)$. $\because f(n-1)\mid f(n)$, if $f(n-1)<f(n)$, then $\exists d\geq2$ s.t. $df(n-1)=f(n)$. If $n$ is not a power of $2$, then $\nu_2(f(n-1))=\nu_2(f(n))$. $\Rightarrow d\geq3$. If $n$ is a power of $2$, then $\nu_2(f(n-1))=\nu_2(f(n))-1$, and $\forall p\geq3$, $\nu_p(f(n-1))=\nu_p(f(n))$. $\Rightarrow d=2$. $\therefore d=2\iff n$ is a power of $2$. $\because x, y>0,\ z>x, z>y$. $\Rightarrow z\geq2x, z\geq2y$. $\Rightarrow x+y=z=\frac{z+z}2\geq x+y$, and the equation holds $\iff z=f(n), x=y=f(n-1)$ for some $n$ which is a power of $2$. $\therefore(x, y, z)=(f(2^k-1), f(2^k-1), f(2^k))$ for $k\in\mathbb Z^+$.

So, taking $v_2$ of superb numbers, we get that this only increases and that this happens only at a power of 2. WLOG $y\geq x$ If $x$ and $y$ have different $v_2$ then their sum has the $v_2$ of the lower. However $z\ge y$ so this is not possible. That means $x$ and $y$ have equal $v_2$. So that adding $x$ and $y$ increases the $v_2$. Now, note that whenever the $v_2$ increases, the number is doubled. This means $z\ge 2y$ so $x=y$ This means that z is the superb number for a power of 2, and x and y are the superb numbers for 1 less than the power of 2.

Let $x = \mathrm{lcm}(1,2,\ldots, a)$, $y = \mathrm{lcm}(1,2,\ldots, b)$, and $z = \mathrm{lcm}(1,2, \ldots, c)$ for $a,b,c$ minimal and WLOG $a < b < c$. First we see that $\nu_2(x)$ is the largest power of $2$ at most $a$, $\nu_2(y)$ is the largest power of $2$ at most $b$, and $\nu_2(z)$ is the largest power of $2$ at most $c$. If there is a power of $2$ between $a$ and $b$, then $\nu_2(x) <\nu_2(y)$ and since $x + y =z$, $\nu_2(x) = \nu_2(z)$, which is absurd. Therefore, there is no power of $2$ between $a$ and $b$. This means that $\nu_2(z)$ is at least $\nu_2(x) + 1$, so there is a power of $2$ between $b$ and $c$. However, we have $z \ge 2y$, so $x = y$ and $z = 2y$. If $2^n$ is the smallest power of $2$ at most $c$, then $2y \mid \mathrm{lcm}(1,2,\ldots, 2^n) \mid z = 2y$, so $z = \mathrm{lcm}(1,2,\ldots, 2^n)$, implying $c = 2^n$. We have $x = \mathrm{lcm}(1,2, \ldots, a) = \frac{\mathrm{lcm} (1,2, \ldots, 2^n)}{2}$, so $x = \mathrm{lcm}(1,2, \ldots, 2^n - 1)$. Hence the answer is $(x,y,z) = ( \mathrm{lcm}(1,2, \ldots, 2^n - 1) , \mathrm{lcm}(1,2, \ldots, 2^n - 1), \mathrm{lcm}(1,2, \ldots, 2^n))$, for any positive integer $n$, which works.

Define $L_k = \text{lcm}(1, 2, \dots, \ell_k)$ where $\ell_k$ is the $k$th number that is a prime or prime power. We can then assume that $x, y, z$ correspond to specific $L_k$. Note that $x, y, z$ work if $x = y = \text{lcm}(1, 2, \dots, 2^k - 1)$, $z = \text{lcm}(1, 2, \dots, 2^k)$. This is also the only solution where two of $x, y, z$ are equal (as $z$ must then WLOG be double $x, y$). Then note that $L_k + L_{k+1} < L_{k+2}$ holds which finishes.

Pretty easy for a Taiwan MO P2 actually. We define the function $f: \mathbb{N} \to \mathbb{N}$ such that $f(n)=\text{lcm}(1,2,\dots,n)$ for all $n\in \mathbb{N}$. The answers are all triples of integers $(f(2^n - 1), f(2^n-1),f(2^n))$ for all positive integers $n$. It’s easy to see that these triples satisfy the given equation. We now show these are the only solutions. It is easy to see that $f$ is non-decreasing, and that for all $m>n$, $f(n) \mid f(m)$ due to the nature of the $\text{lcm}$ function. We start off with the following key claim. Claim : If $(x,y,z)$ is a triple of solutions to the given equation, then we must have $x=y$. Proof : Without loss of generality assume that $x\geq y$. Then, since clearly $z=x+y >x$, we must have $x\mid z$. Thus, $z=kx$ for some positive integer $k$. Now, if $k>2$, \[y=z-x=kx-x=(k-1)x > x \]which is a clear contradiction. Thus, we must have $k=2$ (since $z>x$ clearly) which implies the claim. Now, we are left with finding all integers $m$ for which both $m$ and $2m$ are superb. If we write the sequence of all superb integers (which is known to be decreasing) with all the repetitions removed (if a bunch of consecutive superb numbers are the same we write only one of them down), $m$ and $2m$ must lie next to each other. This is since if $s$ and $r$ are the terms after $x$ in that order in this sequence, we must have $r \geq 2s \geq 4x > 2m$ which is a clear contradiction. Now, in the original sequence, (with repetitions) this means all terms between the $m$ and $2m$ are also either $m$ or $2m$, so we can consider consecutive $p$ and $p+1$ for which $f(p)=m$ and $f(p+1)=2m$. Now, if $f(p+1)=2^kl$ for positive integers $k$ and $l$, it is clear that $l=1$ since if $l>1$, $2^k<p+1$ so $2^k \mid f(p)$ already which is a clear contradiction to the fact that $\nu_2(f(p+1))=\nu_2(f(p))+1$. Thus, $f(p+1)=m=2^k$, which implies that all solutions are indeed as we claimed.

WLOG $x \leq y$ $\because x \leq y \leq z$, it’s easy to prove that $x|y|z$ since $x$, $y$, $z$ are superb Let $y=ax, z=by=abx$ (where $a,b \in \mathbb{Z}^+$) the original equation becomes $x+ax=abx \rightarrow 1=a(b-1)$ so $a=1$, $b=2$ namely $x=y=\frac z2$ so $(x,y,z)=(lcm(1,2,…,2^k-1),lcm(1,2,…,2^k-1),lcm(1,2,…,2^k))$ (where $k$ is any positive integer)

Answer: $\boxed{(x,y,z)=(\text{lcm}(1,2,\dots,2^k-1),\text{lcm}(1,2,\dots,2^k-1),\text{lcm}(1,2,\dots,2^k)), k\in \mathbb{N}}$ Notice that $f(n)=\text{lcm}(1,2,\dots,n)$ is a positive, non decreasing, function of $n$. Notice that $f(n-1)|f(n)$ so when $f(n)$ does increase it does so by a factor of at least $2$. Thus we must have $2x=2y=z$. Since $f(n)$ increases by a factor $2$ only when $n=2^k$ we get the above solutions.