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?

I know this is an old question but I wanted to add my two cents. What you wronte has a closed for. $\sum_{3j=0}^n\binom{n}{3j} = \frac{2^n+m}{3}$  where m  =  2, 1, −1, −2, −1, 1, when n is congruent to
0, 1, 2, 3, 4, 5 (mod 6). according to https://math.hmc.edu/benjamin/wp-content/uploads/sites/5/2019/06/Sums-of-Evenly-Spaced-Binomial-Coefficients.pdf
That said, let us write your special case
$\sum_{j=1}^{n-1} \binom{3n}{3j} = -2 + \sum_{j=1}^{n-1} \binom{3n}{3j}$
Obvoiusly $3n\equiv 0,or 3 mod{6}$. Then m = 2 or -2. Therefore
$\sum_{j=1}^{n-1} \binom{3n}{3j} = -2 +\frac{2^{3n}\pm2}{3} = \frac{2^{3n}\pm2-6}{3} = \frac{2^{3n}-4}{3} or \frac{2^{3n}-8}{3}$
Since $gcd(3,4)=1$ and $4|2^{3n}-4$ and $4|2^{3n}-8$ Then $4|\sum_{j=1}^{n-1} \binom{3n}{3j} |$ .
This gives us even a stronger version of your conclusion
$4|\sum_{j=1}^{n-1} \binom{3n}{3j}$ if n is even  and $8|\sum_{j=1}^{n-1} \binom{n}{3j}$ if n is odd I hope this helped
