5$\cdot$106 $n$ countries, each with 3 representatives, forming $m$ committees $A_{n}(1)$, $A_{n}(2), \cdots, A_{n}(m)$ is called a conference circle if
（1）each committee has $n$ members, each from a different country, one from each;
（2）no two committees have exactly the same members;
(3) $A_{n}(i)$ and $A_{n}(i+1)$ have no common members, $i=1,2, \cdots, m$. It is also agreed that $A_{n}(m+1)=A_{n}(1)$;
(4) if $1<|i-j|<m-1$, then $A_{n}(i)$ and $A_{n}(j)$ have at least 1 common member.
Does there exist a conference circle consisting of 1990 committees from 11 countries?
[Solution] Exists.
Let the circle of $m$ committees composed of $n$ countries be denoted as $(m, n)$, and the meeting composed of all members of committee $A_{n}(i)$ and the $j$-th representative of the $(n+1)$-th country be denoted as $\left(A_{n}(i), j\right)$.

Let $\left\{A_{n}(1), A_{n}(2), \cdots, A_{n}(m)\right\}$ be a $(m, n)$ circle. When $m$ is odd, $\left\{\left(A_{n}(1), 1\right),\left(A_{n}(2), 2\right),\left(A_{n}(3), 1\right), \cdots,\left(A_{n}(m), 1\right),\left(A_{n}(1), 2\right)\right.$, $\left.\left(A_{n}(2), 1\right), \cdots,\left(A_{n}(m), 2\right)\right\}$ is a $(2 m, n+1)$ circle, where the second component alternates between 1 and 2; when $m$ is even, $\left\{\left(A_{n}(1), 3\right),\left(A_{n}(2), 1\right)\right.$, $\left(A_{n}(3), 2\right),\left(A_{n}(4), 1\right),\left(A_{n}(5), 2\right), \cdots,\left(A_{n}(k-2), 1\right),\left(A_{n}(k-1)\right.$, $2),\left(A_{n}(k), 3\right),\left(A_{n}(k-1), 1\right),\left(A_{n}(k-2), 2\right), \cdots,\left(A_{n}(5), 1\right)$, $\left.\left(A_{n}(4), 2\right),\left(A_{n}(3), 1\right),\left(A_{n}(2), 2\right)\right\}$ is a $(2(k-1), n+1)$ circle, where $k$ is an even number not greater than $m$, and except for the second component of $A_{n}(1), A_{n}(k)$ being 3, the second component of the other committees alternates between 1 and 2.

Obviously, $A_{1}(1)=\{1\}, A_{1}(2)=\{2\}, A_{1}(3)=\{3\}$ form a $(3,1)$ circle. Thus, according to the method of constructing new circles given above, we can sequentially obtain the circles $(6, 2),(10,3),(18,4),(34,5),(66,6),(130,7),(258,8),(514,9)$, $(1026,10)$. By taking $k=995$, we can obtain the $(1990,11)$ circle.