(1). If $\lim_{n\to \infty}a_n=a$ then $(a_n)_n$ has a monotone sub-sequence $(a_{f(n)})_n.$ Proof: Let $C=\{n:a_n=a\}.$ Let $P=\{n:a_n>a\}$ and $Q=\{n:a_n<a\}.$
$\quad$ (i). If $C$ is infinite let $f:\Bbb N\to C$ be the unique strictly increasing bijection.
$\quad$ (ii). If $C$ is finite and $P$ is infinite let $f(1)=\min P$ and let $f(n+1)=\min \{j>f(n): a< a_j< a_{f(n)}\}.$
$\quad$ (iii). If $S$ and $P$ are finite let $f(1)=\min Q$ and let $f(n+1)=\min \{j>f(n):a_{f(n)}<a_j<a\}.$
(2). A bounded sequence $(b_n)_n$ has a convergent sub-sequence $(b_{g(n)})_n.$ Proof: Suppose $\{b_n:n\in \Bbb N\}\subset [l,u].$
Let $[l_1,u_1]=[l,u]$ and let $g(1)=1.$
For convenience, for any $n$ let $m_n=(l_n+u_n)/2.$
Now if $\{n:b_n\in [l_n,m_n]\}$ is infinite let $[l_{n+1},u_{n+1}]=[l_n,m_n];$ otherwise let $[l_{n+1},u_{n+1}]=[m_n,u_n].$ And in either case let $g(n+1)=\min \{j>g(n):b_j\in [l_{n+1},u_{n+1}]\}.$
Observe that $|b_{g(n)}-b_{g(n+1)}|\le u_n-l_n=2^{1-n}(u-l)$, which implies that $(b_{g(n)})_n$ is a Cauchy sequence.
(3). Let $b_n=\arctan x_n \in (-\pi/2,\pi/2).$ By (2) there exists a convergent sub-sequence $(b_{g(n)})_n$ and by (1), with $a_n=b_{g(n)},$ the sequence $(b_{g(n)})_n$ has a monotone sub-sequence $(b_{g(f(n))})_n.$ Since $\tan $ is a monotone function on the domain $(-\pi/2,\pi/2),$ therefore $$(\tan b_{g(f(n))})_n=(x_{g(f(n))})_n$$ is monotone.
Let $b_k = 2^{-l_k}$, or $l_k = -\log_2 b_k$. Then
$$ \begin{align}
a_n&=-\frac{1}{n}\sum_{k=1}^{n} \log_2 b_k - \log_2 n \tag{1}\\
&=-\frac{1}{n}\sum_{k=1}^{n} \log_2(k \,b_k) +\frac{1}{n} \sum_{k=1}^{n}\log_2 k - \log_2 n \tag{2}\\
&=\frac{1}{n}\sum_{k=1}^{n} \log_2 \frac{1}{k \,b_k} +O(1) \tag{3}\\
&\ge \log_2 \frac{n}{\sum_{k=1}^{n} k b_k} +O(1) \tag{4}
\end{align}
$$
where in $(2)\to (3)$ we've used $\sum_{k=1}^{n} \log_2(k) = \log_2 (n!) = n \log_2 n - n \log_2 e +O(\log n)$.
And for $(3)\to (4)$ we've used the log-sum inequality.
Let $B = \sum_{k=1}^\infty b_k \le 1$
If $B=1$, then $b_k$ represents a discrete distribution function. In that case we have
$$ \frac{1}{n} \sum_{k=1}^n k \, b_k \to 0$$
as $n \to \infty$ (proof here). Hence $a_n$ in $(4)$ diverges.
This also (a fortiori) applies for $B<1$.
Best Answer
Example 1. (Using the ideas of Greg Martin and S.H.W.)
Let $A=\{(2\times 4^k)^2~|~k\in \{0\}\cup {\mathbb N}\}.$ Define $f:A\rightarrow {\mathbb R}$ by $$f(n)=n^{-1/2},~{\rm so~}f((2\times 4^{i-1})^2)=\frac 1{2^{2i-1}},i\in {\mathbb N}.$$ Then $$\sum_{n\in A}f(n)=\sum_{i=1}^\infty \frac 1{2^{2i-1}}=\frac 2 3$$ and $\lim_{n\rightarrow \infty}nf(n)=\sqrt{n}\rightarrow \infty$ if $n$ stays in $A.$
To extend the above $f$ to a function $f:{\mathbb N}\rightarrow {\mathbb R},$ let $f(n)=2^{-2n}$ for $n\in {\mathbb N}\setminus A.$ And one checks that $$\sum_{n\in {\mathbb N}\setminus A}f(n)=\sum_{n\in {\mathbb N}\setminus A}2^{-2n} < \sum_{n\in {\mathbb N}}2^{-2n}=\frac 1 3, $$ hence $f$ satisfies the two conditions in the original question but $\lim_{n\rightarrow \infty} nf(n)$ does not exist.
Example 2. Let $f:{\mathbb N}\rightarrow A:=\{2^{-i}~|~i\in {\mathbb N}\}$ be a bijection defined by $$f(i)=2^{-i},i\in {\mathbb N}.$$ Then $\sum_{i=1}^\infty f(i)=1.$ This function satisfies the two conditions in the original question with $\lim_{n\rightarrow \infty}nf(n)=0.$ One can do a “surgery” of $f$ to get a bijection $f^*:{\mathbb N}\rightarrow A$ satisfying the two conditions but $\lim_{n\rightarrow \infty}nf^*(n)$ does not exist. To do this, let $$B:=\{2^{2k-1}~|~k\in {\mathbb N}\}~{\rm and~}C:=\{2^{-2k}~|~k\in {\mathbb N}\}$$ and define $$f^*:B\rightarrow C$$ by $$f^*(2^{2k-1})=2^{-2k},k\in {\mathbb N}.$$ Then $$\lim_{n\rightarrow\infty} nf^*(n)=\frac 1 2~{\rm if~}n\in B.\qquad (1)$$ To get an extension for $f^*$ to $f^*:{\mathbb N}\rightarrow A,$ one just defines a natural map $$f^*:{\mathbb N}\setminus B\rightarrow A\setminus C$$ such that ${\mathbb N}\setminus B$ and $A\setminus C$ retain the natural ordering, and $f^*$ maps the $j$th element of ${\mathbb N}\setminus B$ to the $j$th element of $A\setminus C.$ Then $f^*:{\mathbb N}\rightarrow A$ will be as follows if one lists the first few terms, where $2,8,32,$ etc. are elements in $B$.
$$\begin{array} {|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline i & 1 &{\bf 2}& 3& 4& \cdots& 7& {\bf 8}& 9&\cdots& 23& \cdots&{\bf 32}\\ \hline f^*(i) & 2^{-1} & 2^{-2}& 2^{-3}& 2^{-5}& \cdots& 2^{-11}& 2^{-4} & 2^{-13}& \cdots& 2^{-41}& \cdots& 2^{-6}\\ \hline \end{array}$$
By construction, $\sum_i f^*(i)=\sum_if(i)=1$ and $f^*(i)=2^{-l_i}$ for some $l_i\in {\mathbb N}.$ Now one can find a subsequence of ${\mathbb N}$ consisting of prime numbers $p$ such that $$2^{2k-1}<p<2^{2k+1},$$ the existence of which is guaranteed by the Bertrand’s postulate. One can show that $$\lim_{p\rightarrow\infty}pf^*(p)=0.\qquad (2)$$
To see this, suppose $p$ (indexed by $k$) with $2^{2k-1}<p<2^{2k+1}$ has been chosen. Then one has $$f^*(p)=2^{-[2(p-k)-1]},$$ since $f^*(p)$ has to avoid $k$ terms in $C$ (corresponding to the $k$ terms in $B$ as pre-images, namely $2^1,2^3,\cdots,2^{2k-1}$ on which $f^*$ is defined) with other $p-k$ terms in $A\setminus C=\{2^{-(2i-1)}~|~i\in {\mathbb N}\}$ filled.
It follows that $$pf^*(p)=\frac p{2^{2p-2k-1}}=\frac p{2^{2p}}\cdot 2^{2k+1}<\frac {(2^{2k+1})^2}{2^{2p}}=\frac {(2^2\cdot 2^{2k-1})^2}{2^{2p}}<\frac {2^4\cdot p^2}{2^{2p}}\rightarrow 0$$ as $p\rightarrow \infty.$ This proves (2). Combining (1) and (2), one sees that $\lim_{n\rightarrow \infty}nf^*(n)$ does not exist.
Note. There is much freedom to choose $p.$ For example, one can choose $p_1=7$ with $2^{2\cdot 1-1}<7<2^{2\cdot 2-1}$ and $p_2=23$ with $2^{2\cdot 2-1}<23<2^{2\cdot 3-1},$ so from the table, one sees that $f^*(7)=2^{-[2(7-1)-1]}=2^{-11}$ and $f^*(23)=2^{-[2(23-2)-1]}=2^{-41},$ etc.