a possibly stronger type of attack on identity-based encryption
You can probably prove the security against your game from the security in the IND-ID-CCA game of Boneh and Franklin (see http://courses.cs.vt.edu/cs6204/Privacy-Security/Papers/Crypto/IBE-Weil-Pairing.pdf). The idea is to create an adversary $\mathcal{B}$ against IND-ID-CCA from your adversary $\mathcal{A}$. Essentially $\mathcal{B}$ will play man-in-the-middle between the IND-ID-CCA challenger and $\mathcal{A}$.

When $\mathcal{B}$ receives the $2k+1$ identities from $\mathcal{A}$, he chooses one of those at random (let's call it $ID_0$) sends two random values $(x_0,x_0')$ together with $ID_0$ to the challenger and receives an encryption $C_0$ of either $x_0$ or $x'_0$. For each of the other identities $ID_i$, $\mathcal{B}$ encrypts values $x_i$ chosen in such a way that $(x_0,\cdots, x_{2k+1})$ are valid shares for the secret sharing, the ciphertexts are denoted by $C_i$. Then $\mathcal{B}$ returns the ciphertexts to $\mathcal{A}$. [Note that if the challenger has chosen $x_0$ your game is perfectly emulated, otherwise, the shares are invalid]

When $\mathcal{A}$ sends the $k$ identities, if $ID_0$ is in the set $\mathcal{B}$ aborts the communication with $\mathcal{A}$ and returns a random guess to the challenger. Otherwise, $\mathcal{B}$ asks the $k$ keys to the challenger and sends those to $\mathcal{A}$. When $\mathcal{A}$ returns $y$, $\mathcal{B}$ proceeds as follow:
If $y$ is consistent with the share $x_0$, then $\mathcal{B}$ guesses that the challenger returned an encryption of $x_0$, otherwise $\mathcal{B}$ guesses at random.

If $\mathcal{A}$ wins your game with probability $\epsilon$, the advantage of $\mathcal{B}$ is approximately $\epsilon/4$. 



Additional information:


The advantage is not $\epsilon/2$ as I wrote initially but $\epsilon/4$, this is due to the fact that there are two possible definitions for the advantage of a distinguisher. Some papers use the absolute value of the difference to $1/2$ and some papers use twice that. For technical reasons, twice the difference is preferable and I did not initialy noticed that Boneh-Franklin don't use this version.
How is the advantage computed ? This needs an additional hypothesis which I did not mention. Take your game with a variable number of shares, say $\ell$, keeping $k$ fixed and denote by $\ell_0$ the smallest value for which there exists an adversary with non-negligible advantage. Clearly, if there exists an adversary for $\ell=2k+1$ then $\ell_0\leq 2k+1$. And, of course, $\ell_0&gt;k$ because with $k$ shares you have no information about $x$.

If $\ell_0=2k+1$ use the above reasoning. in this case, $\mathcal{A}$ cannot return $y=x$ with non-negligible probability when one share contains $x'_0$ because you could then transform $\mathcal{A}$ into an adversary against $\ell_0-1$ shares by just adding a random valued share. If $\ell_0&lt;2k+1$, redo the reasoning with the new $\ell_0$. This changes the advantage to $\frac{(\ell_0-k)\epsilon}{2k}$. The only problem is that in the case $\ell_0=k+1$, the reduction is not tight.




Response to Ricky's comment on the tightness of the reduction:

I don't see how you get a degradation exponential in $k$. Here is a more precise analysis of the full game if you don't assume that the probability of success jumps from negligible for non-negligible but do a more precise analysis.

Let $\epsilon_\ell$ be the best possible advantage for an adversary that wins your game with $\ell$ queries. Note that $\epsilon_\ell$ form an increasing sequence.  

Moreover, the advantage in your original game is:
$$
\epsilon=\epsilon_{2k+1}=\sum_{i=k+1}^{2k+1}\epsilon_i-\epsilon_{i-1}.
$$
As a consequence, at least one of the differences is bigger than $\epsilon/(k+1)$.

Run $\mathcal{A}$ for an $\ell_0$ such that $\epsilon_{\ell_0}-\epsilon_{\ell_0-1}$ is not too small. The probability that $\mathcal{A}$ wins with $\ell_0$ genuine shares is higher than the probability with a fake share by at least  $\epsilon_{\ell_0}-\epsilon_{\ell_0-1}$, otherwise $\mathcal{A}$ could be used to improve $\epsilon_{\ell_0-1}$.

As a consequence, $\mathcal{B}$ wins with advantage at least $\frac{(\ell_0-k)(\epsilon_{\ell_0}-\epsilon_{\ell_0-1})}{2k}\geq \frac{(\ell_0-k)\epsilon}{2k^2}$.

So the degradation is polynomial in $k$ not exponential.