There's a very pretty combinatorial proof of the general identity $$\sum_{k \geq 0} \binom{n}{rk} = \frac{1}{r} \sum_{j=0}^{r-1} (1+\omega^j)^n,$$
for $\omega$ a primitive $r$th root of unity, in Benjamin, Chen, and Kindred, "Sums of Evenly Spaced Binomial Coefficients," Mathematics Magazine 83 (5), pp. 370-373, December 2010.
They show that both sides count the number of closed $n$-walks on $C_r$ beginning at vertex 0, where $C_r$ is the directed cycle on $r$ elements with the addition of a loop at each vertex, and a walk is closed if it ends where it starts.
Left-hand side: In order for an $n$-walk to be closed, it has to take $kr$ forward moves and $n-kr$ stationary moves for some $k$.
Right-hand side: The number of closed walks starting at vertex $j$ is the same regardless of the choice of $j$, and so it suffices to prove that the total number of closed $n$-walks on $C_r$ is $\sum_{j=0}^{r-1} (1+\omega^j)^n.$ For each $n$-walk with initial vertex $j$, assign each forward step a weight of $\omega^j$ and each stationary step a weight of $1$. Define the weight of an $n$-walk itself to be the product of the weights of the steps in the walk. Thus the sum of the weights of all $n$-walks starting at $j$ is $(1+\omega^j)^n$, and $\sum_{j=0}^{r-1} (1+\omega^j)^n$ gives the total weight of all $n$-walks on $C_r$. The open $n$-walks can then be partitioned into orbits such that the sum of the weights of the walks in each orbit is $0$. Thus the open $n$-walks contribute a total of $0$ to the sum $\sum_{j=0}^{r-1} (1+\omega^j)^n$. Since a closed $n$-walk has weight $1$, $\sum_{j=0}^{r-1} (1+\omega^j)^n$ must therefore give the number of closed $n$-walks on $C_r$.
They then make a slight modification of the argument above to give a combinatorial proof of
$$\sum_{k \geq 0} \binom{n}{a+rk} = \frac{1}{r} \sum_{j=0}^{r-1} \omega^{-ja}(1+\omega^j)^n,$$
where $0 \leq a < r$.
Benjamin and Scott, in "Third and Fourth Binomial Coefficients" (Fibonacci Quarterly, 49 (2), pp. 99-101, May 2011) give different combinatorial arguments for the specific cases you're asking about, $\sum_{k \geq 0} \binom{n}{3k}$ and $\sum_{k \geq 0} \binom{n}{4k}$. I prefer the more general argument above, though, so I'll just leave this one as a link and not summarize it.
Going by the assumption that for an odd $n$ the product of $1, (n-1), (n-a)$ is equal to the sum of the rest of the numbers, we have
$$
1(n-1)(n-a) = \frac{n(n+1)}{2} - 1 -(n-1) - (n-a)
$$
solving this gives $a=\frac{n+1}{2}$
Now for odd $n \ge 5$, we have
$$n-a = \frac{n-1}{2} \ge 2$$
Therefore the numbers $1, n-1, n-a$ are distinct.
A similar assumption for even $n$ with $1, n, (n-a)$ gives
$$
1*n(n-a) = \frac{n(n+1)}{2} - 1 -n - (n-a)
$$
and the solution is $a=\frac{n+2}{2}$
And for even $n \ge 6$ we have
$$n-a = \frac{n-2}{2} \ge 2$$
And therefore the numbers $1, n, n-a$ are distinct.
Best Answer
A generating function solution.
For every $S\subset\{1,2,\ldots,2015\}$ we will write $\Sigma S=\sum_{k\in S}k$.
Let $$ f(a,x) = \prod_{k=1}^{2015} (1+a^kx) = \sum_{S\subset\{1,\ldots,2015\}} a^{\Sigma S} x^{|S|}. $$ Take the average this function over putting $5$th complex roots of unity for $a$. Let $\omega=e^{2\pi i/5}$; then $$ \frac15\sum_{j=0}^4 f(\omega^j,x) = \sum_{S\subset\{1,\ldots,2015\}} \left(\frac15\sum_{j=0}^4 \big(\omega^{\Sigma S}\big)^j\right) x^{|S|} = \sum_{\substack{S\subset\{1,\ldots,2015\}\\\Sigma S\equiv0\pmod5}} x^{|S|}. \tag{$*$} $$ On the RHS of $(*)$, the coefficient of $x^n$ is the number of $n$-element sets $S\subset\{1,\ldots,2015\}$ with $\Sigma S\equiv0\pmod5$.
On the other hand, $$ f(\omega^j,x)= \begin{cases} (1+x)^{2015} & \text{if } j=0 \\ (1+x^5)^{403} & \text{if } j=1,2,3,4 \end{cases} $$ so on the LHS of $(*)$ the coefficient of $x^n$ is: $\frac15\binom{2015}n$ if $n$ is co-prime with $5$, and $\frac15\binom{2015}n+\frac45\binom{403}{n/5}$ if $5|n$.