Most even numbers is a sum $a+b+c+d$ where $a^2+b^2+c^2=d^2$ I have no idea how hard this conjecture is to prove:
Any even number $n\ge 36$ can be written as $n=a+b+c+d$ where $a^2+b^2+c^2=d^2$ and $a,b,c,d\in\mathbb Z^+$.
Small exceptions are $n=2, 4, 6, 10, 12, 14, 20, 26, 34.\,$ Tested for $n\le 10,000$.
A counter-example would be as interesting as a proof.

$n$ as above must be even
$d-c,c+d|a^2+b^2$
 A: If you rewrite your sum of squares as
$$a_4^2-a_3^2=a_1^2+a_2^2$$
Then all you need to do is find a difference of two squares that equals a sum of two squares.
If the sum of two squares is an odd number $b$ with the form $2k+1$ that is
$$a_1^2+a_2^2=b=2k+1$$
Then since any odd number can be trivially written as the difference of two squares we have
$$\left( \frac{b+1}{2}\right)^2-\left( \frac{b-1}{2}\right)^2=b$$
which immediately gives
$$\left( \frac{b+1}{2}\right)^2-\left( \frac{b-1}{2}\right)^2=a_1^2+a_2^2$$
To fit your constraint above $\left( \frac{b+1}{2}\right)^2$ is even and $\left( \frac{b-1}{2}\right)^2$ is odd.
Most sums of two squares have the form $4k+1$ as you will see by adding these in turn
$$(4k_1+1)^2+(4k_2)^2$$
$$(4k_1+1)^2+(4k_2+1)^2$$
$$(4k_1+1)^2+(4k_2+2)^2$$
$$(4k_1+1)^2+(4k_2+3)^2$$
$$(4k_1+2)^2+(4k_2)^2$$
$$(4k_1+2)^2+(4k_2+1)^2$$
$$(4k_1+2)^2+(4k_2+2)^2$$
$$(4k_1+2)^2+(4k_2+3)^2$$
$$(4k_1+3)^2+(4k_2)^2$$
$$(4k_1+3)^2+(4k_2+3)^2$$
Find the results $(\mod 4)$ and see if you can find other sums of the form $a_4^2-a_3^2=a_1^2+a_2^2$
Note Added to help explain why this approach does not work:
I thought the above might help lead to a proof/disproof of the conjecture. Hopefully the comments below will help clarify why this approach does not work.
if $a_1=\left( \frac{u-v}{2}\right)$, $a_2=\left( \frac{u+v}{2}\right)$, $a_3=\left( \frac{w-x}{2}\right)$ and $a_4=\left( \frac{w+x}{2}\right)$ then
$$n=a_1+a_2+a_3+a_4$$ 
$$n=\left( \frac{u-v}{2}\right)+\left( \frac{u+v}{2}\right)+\left( \frac{w-x}{2}\right)+\left( \frac{w+x}{2}\right)$$
$$n=u+w \tag 1$$
and
$$a_4^2=a_1^2+a_2^2+a_3^2$$
$$\left( \frac{w+x}{2}\right)^2=\left( \frac{u+v}{2}\right)^2+\left( \frac{u-v}{2}\right)^2+\left( \frac{w-x}{2}\right)^2$$
$$wx= \left( \frac{u+v}{2}\right)^2+\left( \frac{u-v}{2}\right)^2$$
or
$$2wx=u^2+v^2$$
Substituting for $w$ using (1) gives
$$ n=u+\frac{u^2+v^2}{2x}$$ 
$2x$ being a factor of $u^2+v^2$. 
This result I think shines a little light on why the problem of finding $n$ is not possible using this approach; that is the difficulty in finding a general solution to the factorization of $u^2+v^2$.
A: $$a^2+b^2+c^2=d^2$$
$$a=2ps$$
$$b=2ks$$ 
$$c=s^2-p^2-k^2$$ 
$$d=s^2+p^2+k^2$$
$$2n=2ps+2ks+s^2-p^2-k^2+s^2+p^2+k^2$$ 
$$qt=n=s(p+k+s)$$ 
Lay on multipliers and pick the right.
Or other item.
$$a=2s(p-k)$$
$$b=2s(p+k)$$
$$c=p^2+k^2-2s^2$$
$$d=p^2+k^2+2s^2$$
$$2n=2p^2+2k^2+4ps$$
$$p(p+2s)=n-k^2$$
It remains to try all possible  $k$  . That expression was greater than $0$.
