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Condition: $f:Itomathbb R$ is continuous. For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ such that $$ sum_k |y_k - x_k| < delta$$
implies
$$sum_k |f(y_k) - f(x_k)| < epsilon.$$

Is this condition a necessary or sufficient condition of absolution continuity? Note that the order of the logic identifiers has changed.

A function $f: I to mathbbR$ is absolutely continuous on an interval $I$ if for every $epsilon > 0$ there is a $delta > 0$ such that whenever a finite sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$ satisfies
$$ sum_k |y_k - x_k| < delta$$
then
$$sum_k |f(y_k) - f(x_k)| < epsilon$$

Given any function $f:Itomathbb R$ (not necessarily continuous), the condition: For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ such that $$ sum_k |y_k - x_k| < delta$$
implies
$$sum_k |f(y_k) - f(x_k)| < epsilon$$
is trivially true.

Proof: Given any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, just choose $delta = frac12 sum_k |y_k - x_k|$. Then the condition $ sum_k |y_k - x_k| < delta$ will be false and so the implication "$ sum_k |y_k - x_k| < delta$
implies
$sum_k |f(y_k) - f(x_k)| < epsilon$" will be trivially true.

So this conditionis not sufficient for absolute continuity or even continuity.

Yes, they are equivalent. Suppose you choose $delta$ according to the usual definition of absolute continuity with $epsilon$ repalced by $epsilon /2$. If $(a_k.b_k)$ is a disjoint sequence of interval with total length less than $delta$ then $sumlimits_k=1^N |f(b_k)-f(a_k)| < epsilon /2$ for each $N$. Let $N to infty$ to complete the proof.