For i.i.d. $\{X_n\}$ proving $\frac1n\max\limits_{1\leqslant k\leqslant n}|X_k|\xrightarrow{\mathrm{a.s.}}0\Leftrightarrow E(|X_1|)<+∞$

Hints:

1. Show that $$\frac{1}{n} \max_{1 \leq k \leq n} |X_k(\omega)| \to 0 \tag{3}$$ if, and only if, $$\forall \epsilon>0 \, \, \exists N \in \mathbb{N} \, \, \forall n \geq N: \quad |X_n(\omega)| \leq n \epsilon. \tag{4}$$
2. Use the Borel cantelli lemma to show that $(4)$ (and hence $(3)$) holds almost surely if, and only if, $$\forall \epsilon>0: \quad \sum_{n \geq 1} \underbrace{\mathbb{P}(|X_n| \geq n \epsilon)}_{\mathbb{P}(|X_1| \geq n \epsilon)} < \infty.$$
3. Combine Step 2 with $(2)$ to prove the assertion.

Remark: Step 1 is actually a purely deterministic result. For any sequence $(a_n)_{n \in \mathbb{N}} \subseteq \mathbb{R}$ of real numbers the following equivalence holds: $$\frac{a_n}n \xrightarrow[]{n \to \infty} 0 \iff \frac1n\max_{k \leq n} |a_k| \xrightarrow[]{n \to \infty} 0.$$