When $$$i = 9800$$$, we have 9801 nodes which gives $$$\binom{9801}{2}$$$ possible answers. We split those nodes into 34 groups of at most 290 nodes each. This yields $$$\frac{34 \cdot 35}{2} = 595$$$ possible pairs of groups (note that a pair can consist of the same group twice). We can transmit the correct pair using at most 1 message in the range $$$9800 \le i \lt 9960$$$, because the number of possible message states we can send is $$$1+4 \cdot 160 = 641 \gt 595$$$.

Previously, we narrowed down the answer to 2 groups of at most 290 nodes. We then add 160 new nodes to both groups, giving us 2 groups of $$$290+160=450$$$ nodes. Next, we split each of these into 10 subgroups of 45 nodes, resulting in $$$10^2 = 100$$$ candidate pairs of subgroups. We can communicate the correct pair using at most 1 message in the range $$$9960 \le i \lt 9985$$$ because our available message states is $$$1+4 \cdot 25 = 101 \gt 100$$$.

Following the same logic, we now have 2 groups of $$$45+25=70$$$ nodes. We split each group into 5 subgroups of 14 nodes. The number of unique message states we can send is $$$1+4 \cdot 6 = 25$$$ equal to $$$5^2 = 25$$$, which perfectly matches the $$$5^2=25$$$ pairs of subgroups.

Similarly, we now have 2 groups of $$$14+6=20$$$ nodes. We split each group into 4 subgroups of 5 nodes. The number of unique message states we can send is $$$1+4 \cdot 4 = 16$$$, which perfectly matches the $$$4^2=16$$$ pairs of subgroups.

Continuing the pattern, we have 2 groups of $$$5+4=9$$$ nodes. We split each group into 3 subgroups of at most 4 nodes. The number of unique message states we can send is $$$1+4 \cdot 2 = 9$$$, which perfectly matches the $$$3^2 = 9$$$ pairs of subgroups.

At this stage, we have 2 groups of 4 nodes and 1 new node. This leaves $$$4 \cdot 4 + 4 + 4 = 24$$$ possible remaining answers. Since we can transmit 5 different states, we can divide these candidate answers into 5 groups of at most $$$\lceil \frac{24}{5} \rceil = 5$$$ answers. This is equivalent to partitioning a graph of 24 edges into 5 subgraphs of at most 5 edges each, as shown below: Note that, there is an edge appearing in 2 subgraphs and for each subgraph, there are 2 red nodes, 2 blue nodes, 1 green nodes, 4 edges connecting all pairs of red and blue nodes, and 1 edge connecting the green node to a red node.

At this point, we are left with 5 nodes containing 5 edges, plus 1 new node connected to all 5 existing nodes. We can divide these 10 total remaining edges into 5 groups of $$$\frac{10}{2} = 2$$$ edges each. Note that there is always a central node connecting 2 other nodes.

In this final stage, we have 3 nodes with 2 edges, plus 1 new node connected to the 3 existing nodes. We can easily send a final message to isolate the exact answer among the 5 remaining edges.