Flow25 wrote: Let $n$ be an positive integer. We call $n$ $\textit{good}$ when there exists positive integer $k$ s.t. $n=k(k+1)$. Does there exist 2022 pairwise distinct $\textit{good}$ numbers s.t. their sum is also $\textit{good}$ number? Yes, Choose $2021$ any parwise distinct good numbers. Let $S$ be their sum. Note that each good number is even and so is $S$ Juste write $S+(\frac S2-1)\frac S2=\frac S2(\frac S2+1)$ LHS is a sum of $2022$ good numbers (obviously paitrwise distinct since $(\frac S2-1)\frac S2$ is greater than $S$ and so different of each existing number) RHS is a good number. Q.E.D.

an example can be constructed for this problem or for any natural number in general

This problem was proposed by Burii.

Let $G$ be the set of all good numbers. Obviously we must have that if $n\in G$ then $n$ must be even. Lemma: There are infinitely many triplets $(m,n,g)$ such that $m+n=g$, where $m,n,g \in G$. Proof: If $m,n,g$ are in $G$, then we have that $m=k_1(k_1+1)$, $n=k_2(k_2+1)$ and $g=t(t+1)$. Then by rearranging we have that $m,n,g$ must satisfy: $$k_1(k_1+1)=(t-k_2)(t+k_2-1)$$this implies that if $k_1=ab$ and $k_1+1=cd$, where $(ab,cd)=1$, then $t-k_2=ac$, $t+k_2-1=bd$. Solving for $k_2$, we get that $k_2=\frac{1}{2}(bd-ac-1)$, in order for this to be an integer we must have that $bd \equiv ac+1 \pmod{2}$. Since $k_1$, and $k_1+1$ are of different parity, we have that one of $a,b,c,d$ must be even thus, the modular relation up above holds if we select only one variable to be even, which we choose and we choose so that $bd-ac-2ab-1 > 0$, to get that $k_2 > 0$ and $k_2 > k_1$. Thus there is an infinitely number of solutions. Now back to the problem, let $g_1,g_2,\dots g_n \in G$, choose so that $g_1+g_2$ is good, then we can deduce by the upper lemma that there exists a good number $g_3$, so that $g_1+g_2+g_3$ is also good, choose $g_4$, so that $g_1+g_2+g_3+g_4$ is good,... By induction we choose $g_n$ so that $g_1+g_2+g_3+\dots +g_{n-1}+g_n$ is good, more specifically we choose the good numbers so that $\forall k \leq n$: $$\sum_{i=1}^{k}g_i \in G$$ If we set $n=2022$ we are done.

Let $S_n$ be sum of $n$ good numbers. $S_n = \frac{S_n}{2}((\frac{S_n}{2}+1) - (\frac{S_n}{2}-1)) \implies S_n + \frac{S_n}{2}(\frac{S_n}{2}-1) = \frac{S_n}{2}(\frac{S_n}{2}+1)$ so $\frac{S_n}{2}(\frac{S_n}{2}+1)$ is sum of $n+1$ good numbers Now we just have to check that $\frac{S_n}{2}(\frac{S_n}{2}-1)$ is not in $S_n$ Note that for $S>6$ we have $\frac{S_n}{2}(\frac{S_n}{2}-1) > S_n$ so for $n>6$ we can build $\frac{S_n}{2}(\frac{S_n}{2}+1)$. Let $n = 2021$ and then $\frac{S_{2021}}{2}(\frac{S_{2021}}{2}+1)$ is sum of $2022$ pairwise distinct good numbers.