During a school dance performance, a total of 430 different pairs danced. The first girl danced with 12 different boys, the second girl with 13 different boys, and so on, until the last girl danced with every boy. How many girls and how many boys attended the performance?
If we denote the number of girls by $x$, then the $x$-th girl danced with $x+11$ boys. According to the problem, on one hand, $x+11$ is the total number of boys, and on the other hand,

\[
(1+11)+(2+11)+\ldots+(x+11)=430
\]

The left side is an arithmetic series with $x$ terms, the first term being 12, and the last term being $x+11$, so its sum is

\[
\frac{x}{2}(12+x+11)=430
\]

which simplifies to

\[
x^{2}+23 x-860=0
\]

from which

\[
x_{1}=20 \quad\left[x_{2}=-43\right]
\]

Only the positive root makes sense. Therefore, there were 20 girls and 31 boys at the party.