Divisibility of Sum of Equally Spaced Binomial Coefficients According to a numerical calculation I did for small values of $k$, it appears that the following is true.
$$4|\left[\sum_{j=1}^{n-1}\binom{3n}{3j}\right]$$  or $$\sum_{j=1}^{n-1}\binom{3n}{3j}=4p, p\in\mathbb{Z}$$
Ex. 
If $n=2, \binom{6}{3}=20=4\cdot 5$
If $n=3, \binom{9}{3}+\binom{9}{6}=2\cdot 84=168=4\cdot 42$
If $n=4, \binom{12}{3}+\binom{12}{6}+\binom{12}{9}=2\cdot 220+924=1364=4\cdot341$
If $n=5, \binom{15}{3}+\binom{15}{6}+\binom{15}{9}+\binom{15}{12}=2\cdot 455+2\cdot 5005=10920=4\cdot 2730$
Is there a way to prove this?  Using induction, as above I've shown the base case is true.  Then if we assume that 
$$S_m=\sum_{j=1}^{m-1}\binom{3m}{3j}=4q, q\in\mathbb{Z}$$
Then 
$$S_{m+1}=\sum_{j=1}^{m}\binom{3m+3}{3j}=?$$
And I have no idea how to go forward.  Perhaps its not true?  Is there a counterexample?

Let $$S=\sum_{j=1}^{n-1}\binom {3n}{3j}$$
Adding terms for $j=0$ and $j=n$ gives
$$\begin{align}
S+2
&=\sum_{j=0}^n \binom {3n}{3j}\\
&=\sum_{r=0}^{3n}\binom {3n}r-\left[\sum_{j=0}^{n-1}\binom {3n}{3j+1}+\binom {3n}{3j+2}\right]\\
&=2^{3n}+2\Re\left[\sum_{j=0}^{n-1}\binom {3n}{3j+1}e^{i2\pi/3}+\binom {3n}{3j+2}e^{i4\pi/3}\right]\\
&=8^n+2\Re\left[\sum_{j=0}^{n-1}\binom {3n}{3j+1}e^{i2\pi(3j+1)/3}+\binom {3n}{3j+2}e^{i2\pi(3j+2)/3}\right]\\
&=8^n+2\Re\left[\underbrace{\binom {3n}{3n}+\sum_{j=0}^{n-1}\sum_{k=1}^3\binom {3n}{3j+k}e^{i2\pi(3j+k)/3}}-\sum_{j=0}^{n}\binom {3n}{3j}\right]\\
&=8^n+2\Re\left[\qquad\ \  \quad\overbrace{\sum_{r=0}^{3n}\binom {3n}r e^{i2\pi r/3}}\qquad \qquad -\left(S+2\right)\right]\\
&=8^n+2\Re\big[\big(1+e^{i2\pi /3}\big)^{3n}-\left(S+2\right)\big]\\
&=8^n+2\Re\big[\big(e^{i\pi /3}\big)^{3n}\big]-2\big(S+2\big)\\
&=8^n+2\Re \big[e^{i\pi n}\big]-2\big(S+2\big)\\
&=8^n+2(-1)^n-2\big(S+2\big)\\
3\big(S+2\big)
&=2(-1)^n+8^n\\
3S&=2(-1)^n+8^n-6\\
&\equiv 2(-1)^n+0+2\mod 4\\
&\equiv 2\pm 2\mod 4\\
&\equiv 0\mod 4\\
S=\sum_{j=1}^{n-1}\binom {3n}{3j}&\equiv 0\mod 4\\
\Longrightarrow 4\bigg|&\sum_{j=1}^{n-1}\binom {3n}{3j}\; \blacksquare\end{align}$$
