smc

From the problem, we know that $\frac{1}{3}$,$\frac{1}{4}$,$\frac{1}{6}$, ... are in arithmetic progression. Then, the common difference $d$ of this arithmetic progression is $\frac{1}{4}$ $-$ $\frac{1}{3}$ $=$ $\frac{1}{6}$ $-$ $\frac{1}{4}$ $=$ $-$ $\frac{1}{12}$ Thus, the fourth term of the arithmetic progression is $\frac{1}{12}$. So, the fourth term of the harmonic progression is $12$. Now, we can see that $S_4$ $=$ $3$ $+$ $4$ $+$ $6$ $+$ $12$ $=$ $25$, which corresponds to answer $\boxed{S_4=25}$.