Let $S$ be the final sum, then the necessary conditions are $3|S$, $10|S$.
One can check that for every positive integer $n$ then $S(n) \equiv n \pmod{9}$ which implies that for all ways to insert the plus signs, we have
$$S \equiv 1+1+1+\cdots+1 \equiv 0 \pmod{3} .$$
Then the condition that divisible by $3$ is always satisfied.
Suppose that we inserted $m$ signs in total, then we have $m+1$ parts. Each part is a positive integer that ends with $1$ so the rightmost digit of $S$ is equal to or congruent to $m+1$ modulo $10$. Then we must have $m \in\{9,19,29\}$.
It is easy to see that the given condition is also the sufficiency condition. Denote $a_{1}, a_{2}, \ldots, a_{m+1}$ as the number of digits of each part that partitioned from $m$ signs then
$$a_{1}+a_{2}+a_{3}+\cdots+a_{m+1}=30$$
This equation has $\binom{29}{m}$ solutions. Therefore the number of ways to insert the plus signs is $\binom{29}{9}+\binom{29}{19}+\binom{29}{29}$.