As poncho and Maeher have noted, this isn't possible with straightforward Shamir's secret sharing.
In fact, it's pretty obvious, once you think about it, that there's no way to choose more than $k$ shares independently in advance and get a consistent secret out of them with any unconditionally secure $k$-out-of-$n$ threshold secret sharing scheme, even if you allow the secret to be random.
In particular, imagine you have $k+1$ independently chosen shares $s_1, s_2, \dotsc, s_{k+1}$. By definition, either of the $k$-tuples $(s_1, s_2, \dotsc, s_k)$ and $(s_2, s_3, \dotsc, s_{k+1})$ are enough to fully reconstruct the secret, whereas the $k-1$-tuple $(s_2, s_3, \dotsc, s_k)$ alone provides no information about the share. This means that, for every pair of secrets $S$ and $S'$ and any given $k-1$ shares $s_2, s_3, \dotsc, s_k$, there must exist shares $s_1$ and $s_{k+1}$ such that $(s_1, s_2, \dotsc, s_k)$ yields the secret $S$ while $(s_2, s_3, \dotsc, s_{k+1})$ yields $S'$.
All that said, however, there is a simple way to modify Shamir's secret sharing scheme so that the shares can be chosen independently in advance.
Namely, let $s_1, s_2, \dotsc, s_n$ denote the shares chosen in advance, and let $S$ be the secret you wish to share. Now compute $n$ "auxiliary" shares $a_1, a_2, \dotsc, a_n$ using ordinary Shamir's secret sharing on $S$, and add them to the corresponding pre-chosen shares to get $$p_i := a_i \oplus s_i$$ for all $i \in \{1, 2, \dotsc, n\}$, where $\oplus$ denotes addition in the finite field the shares belong to.
Finally, publish the results $p_1, p_2, \dotsc, p_n$.
Now, to reconstruct the secret $S$, any $k$ participants may each subtract their share $s_i$ from the public $p_i$ (or just reveal their $s_i$ and let others combine it with $p_i$) to obtain the corresponding auxiliary shares $a_i$ and reconstruct $S$ from them.
Meanwhile, assuming that the shares $s_1, s_2, \dotsc, s_n$ are randomly chosen from the finite field used, merely knowing $p_i$ but not $s_i$ for any $i$ provides no information about the corresponding auxiliary share $a_i$. Thus, knowing less than $k$ of the shares will not reveal any information about the secret.
(Of course, if the pre-chosen shares are not uniformly chosen random values, then the unconditional security property will not hold. One may still obtain at least computational security, up to the limits set by brute force guessing attacks, by feeding the pre-chosen shares through a secure cryptographic hash function before combining them with the auxiliary shares.)