Let the number of girls be \( n \). Total matches among girls is given as:
\[ \frac{n(n-1)}{2} = 153 \Rightarrow n(n - 1) = 306 \Rightarrow n = 18 \]
Hence, number of boys in Junior group:
\[ 43 - 18 = 25 \]
Matches between boys and girls in Junior group:
\[ 25 \times 18 = 450 \]
Let the number of boys be \( n \). Matches among boys:
\[ \frac{n(n - 1)}{2} = 276 \Rightarrow n(n - 1) = 552 \Rightarrow n = 24 \]
Hence, number of girls in Senior group:
\[ 51 - 24 = 27 \]
Matches between boys and girls in Senior group:
\[ 27 \times 24 = 648 \]
\[ 450 + 648 = \boxed{1098} \]
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: