Periodicity of Fourier series's coefficients
The reason why Fourier series coefficients of continuous functions are generally not periodic is because of the continuity of the function. The discrete Fourier series coefficients are periodic because the analyzed signals are discrete. Note the duality relationship of the Fourier transform 

$$\text{periodicity}\Longleftrightarrow\text{discreteness} $$

Periodicity in one domain leads to discreteness in the other domain. E.g., a periodic function has only discrete frequency components (as shown by its Fourier series). If these discrete frequency components are periodic, then, consequently, the periodic function must be discrete, i.e., it must be non-zero only at discrete times.

In his answer, Fat32 gave an example of a periodic signal with periodic Fourier series coefficients. The most general form of such a signal is

$$x(t)=\sum_{m=-\infty}^{\infty}b_m\delta\left(t-m\frac{T}{N}\right)\tag{1}$$

where the coefficients $b_m$ are essentially the discrete Fourier series coefficients of the $N$-periodic Fourier coefficients $a_k$ of the $T$-periodic signal $x(t)$:

$$b_m=\frac{T}{N}\sum_{k=0}^{N-1}a_ke^{j2\pi km/N}\tag{2}$$
$$x(t)=\sum_{k=-\infty}^{\infty}a_ke^{j2\pi kt/T}\tag{3}$$

As mentioned above, the signal $x(t)$ is of course discrete in the sense that it is non-zero only at discrete time instances $t_m=mT/N$. Any periodic signal $x(t)$ with $N$-periodic Fourier coefficients $a_k$ must be of the form $(1)$.

A derivation of Equations $(1)$ and $(2)$ can be found in this answer.