Theorem 6.18 in the same chapter of Apostol shows, by partial summation, that
$$L_m(\chi) = L(1,\chi) + O_{\chi}(m^{-1})$$
where the implied constant is effective and depends on $\chi$. This can be explicated by using the Pólya--Vinogradov inequality, which gives
$$L_m(\chi) = L(1,\chi)+O( m^{-1} N^{1/2} \log N)$$
where $N$ is the conductor and the implied constant is effective and absolute (one reference is exercise 11.2.1.3 in p. 374 of Montgomery and Vaughan's book). Taking $m=N$ as you do gives
$$L_N(\chi) = L(1,\chi)+O( N^{-1/2} \log N)$$
unconditionally with an absolute and effective constant. Explicit constants were worked out by various authors, see e.g. Pomerance's paper "Remarks on the Pólya-Vinogradov inequality", Integers 11, No. 4, 531-542, A19 (2011).
To conclude that this is positive, one needs a lower bound on $L(1,\chi)$. There are two lower bounds on $L(1,\chi)$: effective and non-effective.
We have $L(1,\chi)\gg N^{-1/2}$ with an effective constant, see Theorem 11.11 in the Montgomery--Vaughan book, following from a result of Page. This falls just short of being useful here, so I do not think your question can be verified unconditionally.
Siegel showed $$L(1,\chi) > C(\varepsilon)N^{-\varepsilon}$$ for any given $\varepsilon>0$. The positive constant $C(\varepsilon)$ is not effective (it is if one shows there are no Siegel zeros). See Theorem 11.14 of the same book. Taking, say, $\varepsilon=1/3$, we find $L_N(1,\chi)$ is positive for $N$ sufficiently large. Assuming no Siegel zeros, you can make this effective. So conditionally, to verify your question one needs to check finitely many values. I wouldn't be surprised if the answer is positive.
On GRH, $L(1,\chi) \gg 1/\log \log N$ (Littlewood). The constant has been made effective by Lamzouri, Li and Soundararajan in "Conditional bounds for the least quadratic non-residue and related problems" (Math. Comput. 84, No. 295, 2391-2412 (2015); corrigendum ibid. 86, No. 307, 2551-2554 (2017)), see their Theorem 1.5. This should allow you to get a quick verification of your question under GRH.