3. At the high school reunion, former students of one class arrived. All the men shook hands with each other, the women kissed each other (on the cheek), and each man kissed each woman once on the hand. How many people attended the dinner if there were a total of 288 kisses and 78 handshakes?
First method:

Let there be $m$ men. The number of handshakes can be represented by the formula $\frac{m \cdot(m-1)}{2}$.

From the equation $\frac{m \cdot(m-1)}{2}=78$, we get the quadratic equation $m^{2}-m-156=0$.

From the factorization $(m-13) \cdot(m+12)=0$, it is seen that the solutions to the equation are $m_{1}=13$ and $m_{2}=-12$.

Since $m$ must be a natural number, it is concluded that there were 13 men at the dinner.

Let $n$ be the number of women at the dinner. The number of kisses among the women can be expressed as $n \cdot(n-1)$, and the number of kisses between men and women as $13 \cdot n$.

From $n \cdot(n-1)+13 n=288$, after rearranging, we get the quadratic equation $n^{2}+12 n-288=0$.

By factorizing $(n-12) \cdot(n+24)=0$, we arrive at the solutions $n_{1}=12$ and $n_{2}=-24$.

Since $n$ must also be a natural number, it is clear that there were 12 women at the dinner.

In total, there were 25 people at the dinner.