Let $\langle S, \psi\rangle=\sum_{n \in N} \int_0^n \psi'(x)dx$. Is S a distribution?
I claim that S is not a distribution.
I know that if S was a distribution it would satisfy the following property:
$$|\langle S,\psi\rangle|=C_N \sum_{|\alpha \leq N} \|\partial^\alpha \psi\|_\infty$$
So far I have
$$|\langle S,\psi\rangle|=|\sum_{n \in N} \int_0^n \psi'(x)dx| =|\sum_{n \in N} \psi(x) -\psi(0)| =\sum_{n \in N} |\psi(x) -\psi(0)| \leq \sum_{n \in N} |\psi(x)| + |\psi(0)|$$
Your attempt went astray at this step: $$ \left|\sum_{n \epsilon N} \psi(x) -\psi(0)\right| =\sum_{n \epsilon N} |\psi(x) -\psi(0)| \tag{wrong} $$ The triangle inequality gives $\le$, but this does not help you demonstrate your claim (since you wanted to show the sum on the left is large).
Instead, consider a partial sum: $$\sum_{n=1}^M (\psi(x) -\psi(0)) = \sum_{n=1}^M \psi(x) - M \psi(0)$$ and you'll see the root of trouble: it's $M\psi(0)$. The sum $\sum_{n=1}^M \psi(x)$ has a limit as $M\to\infty$, since $\psi$ is a test function. But $M\psi(0)$ does not (unless $\psi(0)$ happens to be zero, which it need not be).