$7 \cdot 74$ There are several locks, and now six people each hold a part of the keys. It is known that any two people trying to open the locks together will have exactly one lock that they cannot open, while any three people can open all the locks. How many locks are there at least?

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[Solution 1] Since every pair of people has one lock they cannot open, let the total number of these locks be $k$, then
$$
k \leqslant C_{6}^{2}=15.
$$

But $k<C_{6}^{2}$ is impossible, because if $k<C_{6}^{2}$, there must be two different pairs of people who cannot open the same lock, and two different pairs of people must include at least three people, which contradicts the fact that any three people can open all the locks.
Therefore, $k=C_{6}^{2}=15$.
That is, there are at least 15 locks.
[Solution 2] Let any pair of people be denoted as $(i, j)$ (numbering the 6 people as $1,2,3,4,5,6$), where $i \neq j$.
Clearly, $(i, j)=(j, i)$.
There are $C_{6}^{2}=15$ such pairs.
Let the lock that $(i, j)$ cannot open be denoted as $a_{ij}$.
We will now prove that when $(i, j) \neq (k, l)$, $a_{ij}$ and $a_{kl}$ are different.
If not, then the pairs $(i, j)$ and $(k, l)$ cannot open the same lock, and $(i, j)$ and $(k, l)$ must include at least three people, which contradicts the fact that any three people can open all the locks.

Therefore, the number of locks that pairs cannot open is the same as the number of pairs, so $C_{6}^{2}=15$ is the minimum number of locks.