This can be proved by application of the 1st Borel-Cantelli lemma which states that: If $(A_n)_{n\in\mathbb{N}}$ is a (not necessarily independent) sequence of events such that $\sum_{n\in\mathbb{N}}\mathbb{P}(A_n)<\infty$, then $\mathbb{P}(A_n\ happens\ infinitely\ often)=0$. We will write $A_n\ i.o.$ to mean $A_n\ happens\ infinitely\ often$.
Now consider that
\begin{align}
limsup\frac{max\{X_1, ..., X_n\}}{n}\leq 1\ \ a.s. & \Leftrightarrow \mathbb{P}(limsup\frac{max\{X_1, ..., X_n\}}{n} > 1)=0 \\
& \Leftrightarrow\mathbb{P}(\forall m>N,\ sup_{n>m}\frac{max\{X_1, ..., X_n\}}{n}>1)=0 \\
& \Leftrightarrow\mathbb{P}(\frac{max\{X_1, ..., X_n\}}{n}>1\ \ \ i.o.)=0
\end{align}
This final implication can be seen from the fact that if $\frac{max\{X_1, ..., X_n\}}{n}>1$ does not happen for infinitely many $n$ (say that the largest $n$ for which this is true is $n'$), then $sup_{n>n'}\frac{max\{X_1, ..., X_n\}}{n}\leq 1$.
Hence applying the Borel-Cantelli lemma, we are left to show that $\sum_{n\in\mathbb{N}}\mathbb{P}(\frac{max\{X_1, ..., X_n\}}{n}>1)<\infty$.
We can calculate that:
\begin{align}
\mathbb{P}(\frac{max\{X_1, ..., X_n\}}{n}>1)&=\mathbb{P}(max\{X_1, ..., X_n\}>n)\\
&=\mathbb{P}(\{X_1>n\}\cup \{X_2>n\}\cup ... \cup \{X_n>n\})\\
&=\mathbb{P}(X_1>n)+\mathbb{P}(X_2>n)+...+\mathbb{P}(X_n>n)\\
&=n\mathbb{P}(X_1>n)
\end{align}
since the $X_i$s are i.i.d
And we see exactly why the conditions were given as they were, since the given condition $\sum_{n\in\mathbb{N}}n\mathbb{P}(X_1>n)<\infty$ implies that $\sum_{n\in\mathbb{N}}\mathbb{P}(\frac{max\{X_1, ..., X_n\}}{n}>1)<\infty$, and so we conclude, by the Borel-Cantelli lemma, that $limsup\frac{max\{X_1, ..., X_n\}}{n}\leq 1$ almost surely