First of all, we notice that $ 11 = 1 + 10, $ and so we write $ 11^n $ as follows: $$(1 + 10)^n = \binom{n}{0} \cdot 1^n + \binom{n}{1} \cdot 1^{n-1} \cdot 10^{1} + \binom{n}{2} \cdot 1^{n-2} \cdot 10^{2} + \cdots$$ We can see that every term after the first two in our expansion has at least two powers of $ 10, $ therefore they will not contribute to the tens digit of anything. Meanwhile, the first term is always $ 1, $ and the second term can be simplified to $ 10n $. Therefore, we have: \begin{align*} &11^1 + 11^2 + 11^3 + \cdots + 11^9 \\ &\qquad\equiv (1 + 10) + (1 + 20) + \cdots + (1 + 90) \pmod{100} \\ &\qquad\equiv 459 \equiv 59 \pmod{100}.\end{align*} Thus, the tens digit must be $ \boxed{5}.$