For any $i < j < k$ we have $n_k \mid S-n_j, S-n_i$ where $S = n_1+n_2+\dots+n_{2022}$; from where it follows that $n_k \mid |n_j-n_i|$. But $|n_i-n_j| < n_j \leqslant n_k$ meaning that $n_i = n_j$ for any three triples $i < j < k$. Thus $n_1 = n_2 = \dots = n_{2021} = x$ (say) and write $n_{2022} = y$, with $y \geqslant x$ and $2022 \in \{x, y\}$. We have from the pair $(1, 2)$ that $x \mid y$. From $(1, 2022)$ we obtain $y \mid 2020x$; hence $y = kx$ with $k \mid 2020$. The first case is that $x = 2022$. In this case we obtain $\tau(2020) = 12$ such sequences. In the other case $y = 2022$ and we have $k \mid 2$ meaning $k = 1$ or $2$ yielding $2$ sequences in this case. Thus overall there are $12+2 = 14$ such sequences. But here, we have counted the constant sequence $\{2022, 2022, \dots, 2022\}$ twice so the answer is $14-1 = 13$.