All-Soviet-Union Mathematical Olympiad

Solution:

Let the coins you suspect to be genuine be $G_1$, $G_2$, ..., $G_7$, and the suspected fakes be $F_1$, $F_2$, ..., $F_7$.

First, weigh $F_1$ against $G_1$. Assuming $F_1$ weighs less, you have proved that $F_1$ is fake and $G_1$ genuine.

Second, weigh $F_1$, $G_2$, $G_3$ against $G_1$, $F_2$, $F_3$. Assuming the first three weigh more, you have proved that they include more genuine coins than the second three. But the second three includes one genuine coin ($G_1$) and the first three includes one fake ($F_1$), so you have proved that $G_2$ and $G_3$ are genuine and $F_2$ and $F_3$ fake.

Finally, weigh $F_1$, $F_2$, $F_3$, $G_4$, $G_5$, $G_6$, $G_7$ against $F_4$, $F_5$, $F_6$, $F_7$, $G_1$, $G_2$, $G_3$. Assuming the first group weighs more, it must include more genuine coins and hence just four genuine coins. Similarly, the second group must include four fakes. So you have proved the identity of the remaining coins.