As $N \to \infty$, the set of fractional parts
$$
\Big\{ \frac{N}{n}\Big\}, \quad 1 \leq n < \sqrt{N} 
$$
becomes asymptotically equidistributed in the Lebesgue measure of $[0,1]$. The same persists with the congruence restriction $n \equiv 1 \mod{2}$ asked by the OP, or indeed with $n$ running through any fixed arithmetic progression. The same equidistribution answer holds moreover in some other natural variants, such as taking $n$ to run through the primes of a fixed primitive arithmetic progression. Regarding this last point I happen to recall de la Vallee Poussin being amused [in an 1898 paper][1] by his observation (*propriete assez curieuse*; *un rapprochement tres remarquable*) that the mean value $1-\gamma$ of fractional parts $\Big\{\frac{N}{an+b} \Big\}$, $an + b \leq N$ was the same for all arithmetic progressions $an+b$, as well as for the prime values $an+b$ when $(a,b) = 1$. (Of course, $ \sum_{n \leq N} \frac{\Lambda(n)}{n} - N^{-1}\log{N!} = \frac{1}{N}\sum_{n \leq N} \Big\{\frac{N}{n}\Big\} \Lambda(n) \sim 1-\gamma$ is just the logarithmic form $\sum_{n \leq N} \frac{\Lambda(n)}{n} = \log{N} - \gamma +o(1)$ of the prime number theorem, [easily equivalent][2] to the usual forms $\psi(N) \sim N$ and $\pi(N) \sim N / \log{N}$.)

Why equidistribution? For simplicity of notation, let me omit the congruence condition on $n$; in practice these are handled straightforwardly as in de la Vallee Poussin's paper, in every situation where we can solve the problem without a congruence condition. It is enough to prove the moment asymptotic
$$
\frac{1}{\sqrt{N}} \sum_{n < \sqrt{N}} \Big\{ \frac{N}{n} \Big\}^k \sim \frac{1}{k+1} = \int_0^1 t^k \, dt, \quad k = 1,2, \ldots,
$$
or what amounts to the same, the estimate
$$
\sum_{n < \sqrt{N}} B_k\Big(\Big\{  \frac{N}{n} \Big\} \Big) = o(\sqrt{N})
$$
for the sum of values of the periodized Bernoulli function $B_k(\{t\})$, for each $k = 1, 2, \ldots$. This is now a more natural form of posing the problem, for there is a conjecture of S. Chowla and H. Walum ("On the divisor problem," *Proc. Symp. Pure Math.*, **vol**. VIII, 1965, pp. 138-143) that asserts the much stronger and best-possible ("square root cancellation") bound
$$
\sum_{n < \sqrt{N}} B_k\Big(\Big\{  \frac{N}{n} \Big\} \Big) \ll_{k,\epsilon} N^{\frac{1}{4} + \epsilon},
$$
for any fixed $k \in \mathbb{N}$ and $\epsilon > 0$. This is one way to generalize the celebrated Dirichlet divisor conjecture, which is the case $k = 1$ with $B_1(\{t\}) = \{t\} - 1/2$. The Chowla-Walum conjecture thus implies, moreover, a quantitative estimate on the discrepancy (rate of equidistribution) of our fractional parts. 

I have not given you the proof of the equidistribution, but I hope this can convince you of the answer to your question and how you may approach it. I would guess that Voronoi's method for the Dirichlet divisor problem could be adapted to prove the crude $o(\sqrt{N})$ bound needed for the qualitative equidistribution. A more precise study of the distribution of fractional parts $\{N / n\}$, $n \leq y$ is done in [these][3] [papers][4] of Saffari and Vaughan, but the proved results are very far from the discrepancy prediction that the Chowla-Walum conjecture implies on your question.

  [1]: https://oeis.org/w/images/a/af/Sur_les_valeurs_moyennes_de_certaines_fonctions_arithm%C3%A9tiques.pdf
  [2]: https://mathoverflow.net/questions/95743/why-could-mertens-not-prove-the-prime-number-theorem/208228#208228
  [3]: http://www.numdam.org/article/AIF_1976__26_4_115_0.pdf
  [4]: http://www.numdam.org/article/AIF_1977__27_2_1_0.pdf