Only 1

Let $f(n)$ as the sum of all digits of a natural number $n.$

Note that $111\cdots1 \equiv 1 \pmod{10}.$

Then if we divide the 1's into $n$ parts, $X\equiv n \pmod{10}.$

Since $X$ is divisible by $30,$ we need $n$ to be $10.$

Nextly, note the strangeness of the multiples of $3.$

"If $\displaystyle N=\sum_{k=1}^n a_k,$ then $\displaystyle f(N)\equiv \sum_{k=1}^n f(a_k) \pmod{3}.$ "

This is because $10^n-10^k\equiv0 \pmod{3}$ for all integers $n\ge1$ and $k\ge0.$

(= Advancing more up doesn't change anything about the modular value of the sum of the digits)

Since $\displaystyle \sum_{k=1}^n a_k$ is always going to be $15$ in this case, $X$ must be a multiple of $3.$

Therefore the number of ways to write those symbols in the squares is equivalent to the number of ways to select $9$ squares and writing $+$ 's in them.

$_{14}C_9=\dfrac{14\cdot13\cdot12\cdot11\cdot10}{5\cdot4\cdot3\cdot2\cdot1}=\boxed{2002}.$