Can we use Shamir secret sharing conversely?
No, if $n$ users have independently chosen secret random numbers $r_i$, and for any given $t\in[1,n)$, we can't use Shamir secret sharing to construct a secret dependent on $t$ out of $n$ of the $r_i$, and independent of which $r_i$ are picked. The critical issue is that what Shamir secret sharing assigns to a user is dependent on what the others get.

More generally, I think there is no function taking $t\in[1,n)$ pairs $(i,r_i)$ with distinct $i$ and producing something independent of the picks but dependent on each $r_i$. I'm trying to think of a proof, but nothing has happened yet.