The sum is 1

For $N=\dfrac{1}{8}$ : If $a=b=c=d=\dfrac{1}{4}$ , then $6a^3+6b^3+6c^3+6d^3= a^2+b^2+c^2+d^2+N$ , so the maximum value of $N$ is at most $\dfrac{1}{8}$ . We will show that for $N=\dfrac{1}{8}$ the inequality holds true.

Using Chebyshey's inequality $4(a^3+b^3+c^3+d^3)\geq (a+b+c+d)(a^2+b^2+c^2+d^2)$ , so $4(a^3+b^3+c^3+d^3)\geq a^2+b^2+c^2+d^2$ . Using Chebyshey's inequality again: $2(a^3+b^3+c^3+d^3)\geq \dfrac{(a+b+c+d)^2}{8}=\dfrac{1}{8}$ By adding the inequalities we get that $6a^3+6b^3+6c^3+6d^3\geq a^2+b^2+c^2+d^2+\dfrac{1}{8}$

Therefore tha answer is: $1+8=\boxed{9}$ .

By the Power Mean Inequality,

$\sqrt[3]{\dfrac{a^3 + b^3 + c^3 + d^3}{4}} \geq \sqrt{\dfrac{a^2 + b^2 + c^2 + d^2}{4}}.$

This gives $4(a^3 + b^3 + c^3 + d^3)^2 \geq (a^2 + b^2 + c^2 + d^2)^3.$ Square rooting both sides and multiplying by three yields

$6(a^3 + b^3 + c^3 + d^3) \geq 3(a^2 + b^2 + c^2 + d^2)^{3/2}.$

Thus, our inequality becomes $3(a^2 + b^2 + c^2 + d^2)^{3/2} \geq a^2 + b^2 + c^2 + d^2 + N.$ Letting $x = a^2 + b^2 + c^2 + d^2$ and subtracting $x$ from both sides of the inequality yields $3x^{3/2} - x \geq N.$

Notice that the maximum possible value of $N$ is precisely the minimum possible value of $3x^{3/2} - x.$ First, we shall find the absolute minimum value of this function without any restrictions on the value of $x.$ Let $f(x) = 3x^{3/2} - x.$ We have $f'(x) = \frac{9}{2}x^{1/2} - 1.$ Setting this equal to 0 gives $x = \frac{4}{81}.$ Since $f''(x) = \frac{9}{4}x^{-1/2} > 0,$ for all positive real values of $x,$ we conclude that at $x = \frac{4}{81},$ the value of $3x^{3/2} - x$ is at its absolute minimum.

However, by the Power Mean Inequality again,

$\begin{aligned} \sqrt{\dfrac{a^2 + b^2 + c^2 + d^2}{4}} &\geq \dfrac{a + b + c + d}{4} \\ a^2 + b^2 + c^2 + d^2 &\geq \dfrac{(a + b + c + d)^2}{4} \\ x &\geq \dfrac{1}{4}. \end{aligned}$

Since $\frac{1}{4} > \frac{4}{81},$ the actual minimum value of $3x^{3/2} - x$ occurs at $x = \frac{1}{4}.$ Thus,

$N \leq 3 \left ( \dfrac{1}{4} \right )^{3/2} - \dfrac{1}{4} = \dfrac{3}{8} - \dfrac{1}{4} = \dfrac{1}{8},$

and $x + y = 1 + 8 = \boxed{9}.$ Equality holds when $a = b = c = d = \frac{1}{4}.$