Junior Balkan Mathematical Olympiad

Solution:
We observe that if $u \leq v$ then by replacing $(u, v)$ with $(u-\varepsilon, v+\varepsilon)$, where $\varepsilon>0$, the sum of squares increases. Indeed,
$$(u-\varepsilon)^{2}+(v+\varepsilon)^{2}-u^{2}-v^{2}=2 \varepsilon(v-u)+2 \varepsilon^{2}>0$$
Then, denoting
$$E(a, b, c, d, x, y, z, t)=a^{2}+b^{2}+c^{2}+d^{2}+x^{2}+y^{2}+z^{2}+t^{2}$$
and assuming without loss of generality that $a \leq b \leq c \leq d$ and $x \leq y \leq z \leq t$, we have
$$\begin{aligned} E(a, b, c, d, x, y, z, t) & \leq E(0,0,0,0, a+x, b+y, c+z, d+t) \\ & \leq E(0,0,0,0,1, b+y, c+z, a+d+x+t-1) \\ & \leq E(0,0,0,0,1,1, c+z, a+b+d+x+y+t-2) \\ & \leq E(0,0,0,0,1,1,1,5)=28 \end{aligned}$$
Note that if $(a, b, c, d, x, y, z, t) \neq(0,0,0,0,1,1,1,5)$, at least one of the above inequalities, obtained by the $\epsilon$ replacement mentioned above, should be a strict inequality. Thus, the maximum value of $E$ is 28, and it is obtained only for $(a, b, c, d, x, y, z, t)=(0,0,0,0,1,1,1,5)$ and permutations of $a, b, c, d$ and of $x, y, z, t$.

Alternative solution by PSC.
Since $0 \leq a, b, c, d \leq 1$ we have that $a^{2} \leq a, b^{2} \leq b, c^{2} \leq c$ and $d^{2} \leq d$. It follows that
$$a^{2}+b^{2}+c^{2}+d^{2} \leq a+b+c+d$$
Moreover, using the fact that $y+z+t \geq 3$, we get that $x \leq 5$. This means that
$$(x-1)(x-5) \leq 0 \Longleftrightarrow x^{2} \leq 6 x-5$$
Similarly we prove that $y^{2} \leq 6 y-5, z^{2} \leq 6 z-5$ and $t^{2} \leq 6 t-5$. Adding them we get
$$x^{2}+y^{2}+z^{2}+t^{2} \leq 6(x+y+z+t)-20$$
Adding (1) and (2) we have that
$$\begin{aligned} a^{2}+b^{2}+c^{2}+d^{2}+x^{2}+y^{2}+z^{2}+t^{2} & \leq a+b+c+d+6(x+y+z+t)-20 \\ & \leq 6(a+b+c+d+x+y+z+t)-20=28 \end{aligned}$$
We can readily check that the equality holds if and only if $(a, b, c, d, x, y, z, t)=(0,0,0,0,1,1,1,5)$ and permutations of $a, b, c, d$ and of $x, y, z, t$.