Entropy change when indistinguishable particles suddenly become distinguishable We have that 

$$
S = \frac{U}{T} + k\ln Q
$$

Using that for indistinguishable particles in the canonical ensemble

$$
\ln Q_{\mathrm{indis}} = N\ln \frac{q_{\mathrm{tr}}e}{N} + N\ln q_{\mathrm{rot}} + N\ln q_{\mathrm{vib}} + N\ln q_{\mathrm{el}}
$$

Similarly for distinguishable particles we have that

$$
\ln Q_{\mathrm{dis}} = N\ln q_{\mathrm{tr}} + N\ln q_{\mathrm{rot}} + N\ln q_{\mathrm{vib}} + N\ln q_{\mathrm{el}}
$$

Then we assume the indistinguishable particles suddenly become distinguishable. By how much does the entropy of the system change?

$$
\Delta S = S_{\mathrm{dis}} - S_{\mathrm{indis}} = k(\ln Q_\mathrm{dis} - \ln Q_\mathrm{indis}) = Nk[\ln(N)-1]
$$

In the special case that the system contains only one particle ($N=1$), the entropy changes by $-k$.

Intuitively, it does not matter whether one particle becomes distinguishable from the rest if there is only one particle; Assigning a label to just one particle does not really modify the information within the system. 

So how can this be understood?
Very interesting question. The issue is that your formula for $\ln Q_\mathrm{indis}$ does not hold for $N = 1$.

Since the rotational, vibrational and electronic degrees of freedom do not come into play I will just ignore them. The way you derived the term $\ln (q_\mathrm{tr}e/N)$ comes from the use of Stirling's approximation

$$\begin{align}
Q_\mathrm{tr,indis} &amp;= \frac{q_\mathrm{tr}^N}{N!} \\
\ln Q_\mathrm{tr,indis} &amp;= N \ln q_\mathrm{tr} - \ln(N!) \\
&amp;\approx N \ln q_\mathrm{tr} - N \ln N + N \\
&amp;= N\ln\left(\frac{q_\mathrm{tr}e}{N}\right)
\end{align}$$

Plugging $N = 1$ into Stirling's approximation however tells you

$$\ln (1!) \approx 1\ln 1 - 1$$

or $0 \approx -1$, which is obviously a problem. Stirling's approximation is really meant to be used for large $N$ only!

Now, you could try working it out without using the approximation: plug $N = 1$ into the line before the approximation $\ln Q_\mathrm{tr,indis} = N \ln q_\mathrm{tr} - \ln(N!)$ and you will get the result

$$\ln Q_\mathrm{tr,indis} = \ln q_\mathrm{tr}$$

which matches the expression for the distinguishable case, so all your terms will cancel out and you will get the desired answer of $\Delta S = 0$.