For any real number $$t$$, let $$\lfloor t\rfloor$$ denote the largest integer $$\leq t$$. Suppose that $$N$$ is the greatest integer such that $$\left\lfloor\sqrt{\left\lfloor\sqrt{\left\lfloor\sqrt{N}\right\rfloor}\right\rfloor}\right\rfloor=4$$. Find the sum of digits of $$N$$.

Let
$$a_1=\left\lfloor\sqrt{N}\right\rfloor,\qquad a_2=\left\lfloor\sqrt{a_1}\right\rfloor,\qquad a_3=\left\lfloor\sqrt{a_2}\right\rfloor.$$

The given condition is $$a_3=4.$$ Starting from the outermost floor and working inward:

Case for $$a_3=4$$
Since $$a_3=\left\lfloor\sqrt{a_2}\right\rfloor=4,$$ we must have
$$4\le \sqrt{a_2}\lt 5\; \Longrightarrow\; 16\le a_2\lt 25.$$

The largest integer satisfying this is $$a_2=24.$$ (Choosing any smaller value will yield a smaller final $$N$$, and $$a_2=25$$ is not allowed because it makes $$a_3=5$$.)

Now impose $$a_2=24$$ on the previous step: $$a_2=\left\lfloor\sqrt{a_1}\right\rfloor=24 \;\Longrightarrow\; 24\le\sqrt{a_1}\lt 25 \;\Longrightarrow\; 576\le a_1\lt 625.$$

The largest integer in this interval is $$a_1=624.$$

Finally impose $$a_1=624$$ on the innermost step: $$a_1=\left\lfloor\sqrt{N}\right\rfloor=624 \;\Longrightarrow\; 624\le\sqrt{N}\lt 625 \;\Longrightarrow\; 624^{2}\le N\lt 625^{2}.$$

Compute the bounds:
$$624^{2}=389,376,\qquad 625^{2}=390,625.$$

Hence the greatest integer $$N$$ satisfying all the conditions is
$$N_{\max}=390,625-1=390,624.$$

Sum of the digits of $$390,624$$ is
$$3+9+0+6+2+4=24.$$

Therefore, the required answer is
24.