There appears to be no mention in OEIS: Quarter-squares, A002620. Can someone give a proof or reference?
Examples:
quarter-squares: ${0,1,2,4,6,9,12,16,20,25,30,36,...}$
2-term sums: ${2+1, 4+1, 6+2, 6+2,...,90+9,...}$
3-term sums: ${12+2+1, 16+2+1,...,72+6+2,...}$
4-term sums: ${240+12+2+1,...,6480+72+6+2,...}$
The quarter-squares are the numbers of form $k^2$ and $k(k+1)$.
Start by expressing $N$ as the sum of four squares.
If you used some square four times, i.e. $N=x^2 + x^2 + x^2 + x^2$, then $N=(2x)^2$ is a quarter-square.
If you used some square three times but not four times, i.e. $N=x^2 + x^2 + x^2 + y^2$, then $N=(x-1)x + x^2 + x(x+1) + y^2$ is a sum of four distinct quarter-squares.
If you used some square twice combined with two other distinct squares, i.e. $N=x^2 + x^2 + y^2 + z^2$, then $N=(x-1)x + x(x+1) + y^2 + z^2$ is a sum of four distinct quarter-squares.
If you used two distinct squares twice each, i.e. $N=x^2 + x^2 + y^2 + y^2$, then $N=(x+y)^2 + (x-y)^2$ is a sum of two distinct quarter-squares.
Finally, if you used four distinct squares, then $N$ is clearly the sum of four distinct quarter-squares.
The quarter-squares are defined by $Q(a) = \lfloor a/2\rfloor\lceil a/2\rceil$. So $Q(2a) = a^2$.
First, write every number as a sum of four squares. We can assume they're not all the same (by induction: if this is a problem, the sum is a multiple of 4; we can produce a better representation by representing n/4 as a sum of four squares not all the same and multiplying everything by $2^2$).
Now, if two or three are the same, then replace two using the identity $a^2 + a^2 = Q(2a+1) + Q(2a-1)$. The only situation where this fails is where our expression is of the form $a^2 + a^2 + (a+1)^2 + (a+1)^2$ (since we'd use $Q(2a+1)$ twice). But this is $4a^2 + 4a + 2$, which is $(2a+1)^2 + 1^2$, so we have an alternative here too.
PS. I notice that Emil has suggested an alternative way of dealing with repeats, while this was being edited into existence.
