We can choose $ M=\{t,2t,3t,...,1992t\}$ in such a way that all numbers $ t,2t,...,\frac{1992(1992+1)}{2}t$ are all perfect powers. You may use Chinese theorem in order to prove it.

Well, take a space containing 1992 line segments of distinct length, and construct a 1992-hedron using these segments or the sum of the lengths of these segments.Now, statement (ii) will imply that there exists a polyhedron which is similar to our polyhedron. But this is impossible, which can easily be shown. Is my solution correct?! : [/quote]

Just reposting my comment. Its fortified with LaTeX now! Well, take a space containing $ 1992$ line segments of distinct length, and construct a $ 1992$-hedron using these segments or the sum of the lengths of these segments.Now, statement (ii) will imply that there exists a polyhedron which is similar to our polyhedron. But this is impossible, which can easily be shown. Is my solution correct?!

Yes. We will take \[ M = \left\{ n, 2n, 3n, \dots, 1992n \right\} \]for some positive integer $n$. It suffices to ensure that $kn$ is a perfect power for each $1 \le k \le C \coloneqq 1+2+\dots+1999$. Let $p_1 < p_2 < \dots < p_m < C$ be the primes less than $C$. Let $q_1 < q_2 < \dots < q_C$ be distinct primes. By the Chinese remainder theorem, for each $1 \le k \le C$, we can choose $\nu_{p_i}(n)$ such that \[ \nu_{p_i}(n) + \nu_{p_i}(k) \equiv 0 \pmod{p_k} \qquad i = 1, \dots, m \]This ensures $kn$ is a perfect $q_k$'th power as needed.