The first Borel-Cantelli lemma yields
$$\mathbb P\left(\limsup_{n\to\infty}|X_n|>n\right)=0. $$
As for each $n$ $$\{|X_n|>n\}\subset \bigcup_{k=n}^\infty \{|X_k|>k\}, $$
it follows that
\begin{align}
0&=\mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty \{|X_k|>k\}\right) \\&=\mathbb P\left(\lim_{n\to\infty}\bigcup_{k=n}^\infty \{|X_k|>k\} \right)\\
&=\lim_{n\to\infty}\mathbb P\left(\bigcup_{k=n}^\infty \{|X_k|>k\} \right)\\
&\geqslant\lim_{n\to\infty}\mathbb P(|X_n|>n)
\end{align}
and hence $\lim_{n\to\infty}\mathbb P(|X_n|>n)=0$. Further, $$\bigcap_{k=n}^\infty \{|X_k|>k\}\subset\{|X_n|>n\} $$
so that
\begin{align}
\mathbb P\left(\limsup_{n\to\infty} |X_n|\leqslant n\right) &= \mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty\{|X_k|\leqslant k\}\right)\\
&=1 - \mathbb P\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty \{|X_k|>k\} \right)\\
&=1 - \mathbb P\left(\lim_{n\to\infty} \bigcap_{k=n}^\infty\{|X_k|>k\} \right)\\
&=1 - \lim_{n\to\infty}\mathbb P\left(\bigcap_{k=n}^\infty\{|X_k|>k\}\right)\\
&\geqslant1 - \lim_{n\to\infty}\mathbb P(|X_n|>n)\\
&= 1,
\end{align}
which implies that
$$\mathbb P\left(\limsup_{n\to\infty} \frac{|X_n|}n\leqslant 1 \right)=1. $$
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
Best Answer
Note that for a non-negative random variable $Y$, we have $$E(Y) = \int_0^\infty P(Y > y) dy$$ Since $S(y) = P(Y > y) = 1 - F_Y(y)$ is a decreasing function in $y$, we have the following Riemann sum approximation: $$E(Y) = \int_0^\infty P(Y > y) dy \leq \sum_{n=1}^\infty P(Y \geq n)$$ Define the events $\Lambda_n$ as you did: i.e. $\Lambda_n = \{ |X_n| \geq a \sqrt{n} \} = \left\lbrace \left(\frac{X_n}{a} \right)^2 \geq n\right\rbrace$. Then, $$\begin{align*} \sum_{n=1}^\infty P(\Lambda_n) &= \sum_{n=1}^\infty P\left(\left\lbrace \left(\frac{X_n}{a} \right)^2 \geq n\right\rbrace \right)\\ &= \sum_{n=1}^\infty P\left(\left\lbrace \left(\frac{X_1}{a} \right)^2 \geq n\right\rbrace \right) \\ &\geq \int P(Y >y)dy & \text{where }Y = \frac{X_1^2}{a^2} \\ &= E(Y) \\ &= \infty \end{align*}$$ Since the events $\Lambda_n$ are independent, the second Borel-Cantelli lemma allows you to conclude that $P\left( \limsup_{n \to \infty} \Lambda_n \right) = 1$.