$4 \cdot 200$ A dance party has 42 participants. Lady $A_{1}$ danced with 7 male partners, Lady $A_{2}$ danced with 8 male partners $\cdots \cdots$ Lady $A_{n}$ danced with all male partners. How many ladies and male partners are there at the dance party?
[Solution] Let there be $n$ ladies at the ball, so the number of male partners is $42-n$. The $k$-th lady $(1 \leqslant k \leqslant n)$ has danced with $k+6$ male partners. Therefore, the $n$-th lady has danced with $n+6$ male partners. According to the given condition, the number of male partners is $n+6$. Thus, $n+6=42-n$. Solving this equation gives $n=18$. Therefore, there are 18 ladies and $42-18=24$ male partners at the ball.