As you noted, if $a_n=\frac{1}{n}$, then $\sum_{n=1}^\infty \frac{a_n}{1+na_n}=\sum_{n=1}^\infty \frac{1}{2n}$ diverges.
On the other hand, say $a_{2^k}=\frac{1}{2}$ and $a_{n}=2^{-n}$ for $n$ not a power of two. Then $\sum_{n=1}^\infty \frac{a_n}{1+na_n} < \sum_{n=1}^\infty \frac{2^{-n}}{1+n2^{-n}} + \sum_{k=1}^\infty \frac{1}{2+2^k} \, ;$ since both right-hand sums are convergent and every term of the left-hand sum is positive, the left-hand sum also converges.
In both cases, $\sum a_n$ diverges. So merely knowing that $\sum a_n$ diverges is insufficient to tell you about the convergence behavior of $\sum \frac{a_n}{1+na_n}$.
On the other hand, if $a_n$ is decreasing and $\sum a_n$ is divergent, then $\frac{a_n}{1+na_n}>\frac{a_n}{1+s_n}>\frac{a_n}{2s_n}$ for $n$ sufficiently large, where $s_n=\sum_{k=1}^n a_k$. Moreover, I claim that if $\sum a_n$ diverges, then so does $\sum \frac{a_n}{s_n}$. Hence, if you add the condition that $a_n$ is decreasing to the original problem, it will in fact be true that $\sum a_n$ converges if and only if $\sum \frac{a_n}{1+na_n}$ does.
To establish the claim, fix some positive integer $N$ and look at the tail of $\sum \frac{a_n}{s_n}$ beginning at the $N^{\rm th}$ term. For any positive integer $k$, we have $\sum_{i=1}^k \frac{a_{N+i}}{s_{N+i}} > \sum_{i=1}^k \frac{a_{N+i}}{s_{N+k}}=\frac{s_{N+k}-s_N}{s_{N+k}}=1-\frac{s_N}{s_{N+k}} \, .$ Since the sequence $s_n$ is divergent, we can choose $k$ such that $s_{N+k}>2s_N$, and thus $1-\frac{s_N}{s_{N+k}}>\frac{1}{2}$. But $N$ was arbitrary; hence the tails of $\sum \frac{a_n}{s_n}$ don't decrease to zero, and so the sum does not converge.