How many positive integers $$𝑛 ≤ 100$$ are divisible by all positive integers $$𝑖$$ such that $$i^{3} ≤ 𝑛 $$?

Let $$n$$ be a positive integer not exceeding $$100$$.
For every such $$n$$ consider the set of divisors that are demanded by the question:

All positive integers $$i$$ with $$i^{3}\le n \quad\Longleftrightarrow\quad 1\le i\le\sqrt[3]{n}\,.$$

Denote $$k=\left\lfloor\sqrt[3]{n}\right\rfloor$$ (the greatest integer whose cube is at most $$n$$). Then $$n$$ must be divisible by each of $$1,2,\dots ,k$$, i.e. by their least common multiple

$$L_k=\operatorname{lcm}(1,2,\dots ,k).$$

Because $$n\le100$$, $$k$$ can only be $$1,2,3,4$$ (since $$5^{3}=125>100$$). We examine each possible $$k$$.

Range of $$n$$: $$1^{3}\le n\le2^{3}-1\;\Longrightarrow\;1\le n\le7$$.
$$L_1=\operatorname{lcm}(1)=1$$, so every $$n$$ in this range works. Count = $$7$$.

Range of $$n$$: $$2^{3}=8\le n\le3^{3}-1=26$$.
$$L_2=\operatorname{lcm}(1,2)=2$$, so $$n$$ must be even.
Even numbers between $$8$$ and $$26$$: $$8,10,12,14,16,18,20,22,24,26$$  ⇒  Count = $$10$$.

Range of $$n$$: $$3^{3}=27\le n\le4^{3}-1=63$$.
$$L_3=\operatorname{lcm}(1,2,3)=6$$, so $$n$$ must be a multiple of $$6$$.
Multiples of $$6$$ in $$[27,63]$$: $$30,36,42,48,54,60$$  ⇒  Count = $$6$$.

Range of $$n$$: $$4^{3}=64\le n\le100$$ (upper bound of the problem).
$$L_4=\operatorname{lcm}(1,2,3,4)=12$$, so $$n$$ must be a multiple of $$12$$.
Multiples of $$12$$ in $$[64,100]$$: $$72,84,96$$  ⇒  Count = $$3$$.

Adding the counts from all cases:
$$7+10+6+3 = 26.$$

Hence, the number of positive integers $$n\le100$$ that satisfy the given condition is 26.