A trivial example showing that this is possible, at least in some cases, is the $n$-out-of-$n$ secret sharing scheme based on modular addition. Let $s \in \mathbb Z / m \mathbb Z$ be the secret, and construct $n$ shares of it by picking $x_1, \dotsc, x_{n-1}$ randomly from $\mathbb Z / m \mathbb Z$ and letting $x_n = s - (x_1 + \dotsm + x_{n-1}) \mod m$. Thus, the secret can be reconstructed by calculating $s = x_1 + \dotsm + x_n \mod m$.
In this scheme, anyone who holds a share $x_i$ can further split their share into $j$ subshares $\xi_1, \dotsc, \xi_j$ in the same way, such that $x_i = \xi_1 + \dotsm + \xi_j \mod m$. If they then discard the original share $x_i$, they'll have expanded the number of shares from $n$ to $n+j-1$. Further, this expansion is completely transparent to the other participants, in that reconstructing the original secret still requires merely adding up all the $n+j-1$ shares modulo $m$.
I can't right now think of any obvious way to devise an $k$-out-of-$n$ secret sharing scheme that could be similarly expanded into an $(k+j)$-out-of-$(n+j)$ scheme by some subset of less than $k$ participants, but I wouldn't be surprised if one did exist.
Addendum: Now that I'm not quite as tired as I was when I first wrote the answer above, I see that the trick I used does not generalize the way the OP apparently wants. In particular, we can prove the following:
Lemma 1: It is not possible to expand an effective $(k,n)$ threshold secret sharing scheme into an effective $(j,m)$ threshold scheme, where $m-j > n-k$, without access to at least $k$ shares.
Proof: Already given by Dilip Sarwate. Essentially, if the holders of $k-1$ shares could do this, they could assign all the $m-n$ new shares to themselves, and so obtain $k+m-n-1 \ge j$ new shares, which would let them recover the secret under the expanded scheme and thus break the original scheme.
Lemma 2: It is not possible to expand an effective $(k,n)$ threshold secret sharing scheme into an effective $(j,m)$ threshold scheme, where $j > k$, without access to at least $n-k+1$ shares.
Proof: As above, if the holders of $n-k$ shares could do this, then the holders of the remaining $k$ shares could still recover the secret, thus breaking the new scheme.
Put together, these lemmata yield the following theorem:
Theorem: It is not possible to expand an effective $(k,n)$ threshold secret sharing scheme into an effective $(j,m)$ threshold scheme, where $m > n$, without access to at least $k$ or $n-k+1$ shares.
For lemma 1, the lower bound of $k$ shares is tight, as shown by poncho. For lemma 2, my example above shows the tightness of the lower bound for the specific case of $k=n$; I'm not sure whether or not it can be tightened further for $k < n$.
Of course, if you're willing to allow more general secret sharing schemes, where not all shares are equivalent, then various kinds of expansion are indeed possible. In particular, it's always possible for any shareholder(s) to further share their own shares with any number of people using any secret sharing scheme of their choosing. These derived shares will not, however, generally be equivalent to the original ones.