2. (Addition to the problem $1^{*}$ ) $n$ countries, each country's team of 3 representatives, form $m$ representative meetings $A_{0}(1), A_{0}(2), \cdots, A_{0}(m)$ called a meeting circle. If
1 each meeting has $n$ representatives, exactly 1,
() no two meetings have the same representatives;
(3) $A_{0}(i)$ and $A_{0}(i+1)$ have no common representatives, $i=1,2, \cdots, m$. It is agreed that $A_{\mathrm{n}}(n l+1)=A_{\mathrm{o}}(1)$,
(1) if $1<|i-j|<m-1$, then $A_{\mathrm{D}}$ (1) and $A_{0}$ (j) have at least one common representative.

Is there a meeting circle of 11 countries with 1990 meetings?
We use $(m, n)$ to denote a conference circle with $n$ countries and $m$ meetings, and use $\left(A_{\mathrm{n}}(i), j\right)$ to denote a concept formed by all the representatives and the $j$-th representative of the $n+1$-th country in the meeting $A_{n}(i)$. If $A,(0), \cdots, A \sim(m)$ is an $(m, n)$ assembly, then when $m$ is an even number, $(4 n(1), 1), 4 n(2), 2), \cdots$, $(A 0(m), 1),\left(A_{0}(1), 2\right),\left(A_{0}(2), 1\right), \cdots$ $(A n(m), 2)$ is a $(2 m, n+1)$ conference circle. In this case, $\left(A_{\mathrm{a}}(1), 3\right),\left(A_{\mathrm{a}}(2), 1\right),\left(A_{\mathrm{a}}(3), 2\right)$, $\left(A_{n}(4), 1\right),\left(A_{0}(5), 2\right), \cdots,\left(A_{n}(k-2), 1\right.$, ; $\left(A_{n}(k-1), 2\right),\left(A_{0}(k), 3\right),\left(A_{n}(k-1), 1\right)$, $(A n(k-2), 2), \cdots,\left(A_{0}(5), 1\right),\left(A_{n}(4), 2\right)$, $\left(A_{n}(3), 1\right),\left(A_{n}(2), 2\right)$. . . a $(2(k-1), 1)$ conference circle, which satisfies an even number not greater than $m$, and except for the first sub-concept corresponding to $A_{n}(1), A_{0}(k)$ modulo 3, all other corresponding second sub-concepts are updated according to the method of constructing new conference circles. Thus, we can sequentially obtain the full circles (6,2), (10,3), (18,4), (34,5), (66, 6), $(130,7),(258,8),(514,9),(1026$, 10), and by taking $k=996$, we get the conference circle $(1990,11)$.