We could use long division to find the decimal representations of the three fractions, but there's a slicker way. We begin by finding an equivalent fraction whose denominator is 1 less than a power of 10. Take $ \frac{3}{11} $, for example. We can multiply the numerator and denominator by 9 to rewrite this number as $ \frac{27}{99} $. Now, we can rewrite this fraction as $ 0.\overline{27} $. To see why, let $ x=0.\overline{27} $, and subtract $ x $ from $ 100x $: $$\begin{array}{r r c r@{}l} &100x &=& 27&.272727\ldots \\ - &x &=& 0&.272727\ldots \\ \hline &99x &=& 27 & \end{array}$$ This shows that $ 0.\overline{27} = \frac{27}{99} $. We can apply the same trick to our other fractions. For $ \frac{4}{37} $, we have to recognize that $ 37\cdot 27 = 999 $, allowing us to write $ \frac{4}{37} $ as $ \frac{4\cdot 27}{37\cdot 27} = \frac{108}{999} $. Now the trick above yields $ \frac{4}{37} = 0.\overline{108} $. To deal with $ \frac{23}{9} $, we first write it as $ 2+\frac{5}{9} $. The trick we used for the other two fractions then gives $ \frac{23}{9} = 2+0.\overline{5} = 2.\overline{5} $. Finally, we find the first six digits after the decimal point of the sum. $$ \begin{array}{c@{}c@{\;}c@{}c@{}c@{}c@{}c@{}c@{}c@{}c}& & 2.&\stackrel{1}{5} & \stackrel{1}{5} & \stackrel{1}{5} & 5 & \stackrel{2}{5} & 5\\& & 0.&2 &7 & 2 & 7& 2 & 7\\&+ & 0.& 1 &0 & 8 & 1 & 0 & 8\\ \hline & &2.& 9 &3 & 6 & 3 & 9 & 0\\ \end{array} $$ We should check that in adding the seventh digits after the decimal point, nothing is carried over to affect the sixth digit. Notice that continuing the addition past the first six digits will result in repeating blocks of the same six digits ($ .555555+.272727+.108108=.936390 $). That means the seventh digit will be a 9 (same as the first digit after the decimal point) and there is nothing carried over to affect the sixth digit. So, the sum $ a+b+c+d+e+f $ is $ 9+3+6+3+9+0=\boxed{30} $.