Is wave spectrum always positive and why?
Yes, wave variance or energy spectrum, direcional or non-directional is positive-definite as @aretxabaleta said in the comment. 

In linear water-wave theory, the surface elevation is described as a linear superposition of sinusoids:

$$
\eta(t) = \sum_{i-1}^{N}a_i \sin(f_i t + \phi_i)
$$

where $a_i$, $f_i$ and $\phi_i$ are the amplitude, frequency and phase, respectively, of each wave component $i$.

The most commonly used wave spectrum is the wave variance spectrum. Wave variance is:

$$
\langle\eta^2\rangle = \dfrac{1}{2N}\sum_{i=1}^{N}a_i^2 = \sigma^2
$$

and wave variance spectrum $F(f)$ is defined such that:

$$
F(f)\Delta{f}=\dfrac{a_i^2}{2} 
$$

In the limit of $N \rightarrow \infty$ (continuous spectrum), the following holds:

$$
\int_{0}^{\infty}F(f)df = \sigma^2
$$

Being quadratic, both wave variance (spectrum integral) and individual discrete spectrum components are positive-definite.

Note that so far we implied non-directional frequency spectrum, i.e. spectrum defined in frequency space. It can be also defined in wavenumber $k$ space, and the following holds:

$$
\int_{0}^{\infty}F(k)dk = \int_{0}^{\infty}F(f)df = \sigma^2
$$

$$
F(k)\Delta{k} = F(f)\Delta{f}
$$

$$
F(k) = F(f)c_g
$$

where $c_g$ is group velocity of an individual component.

The non-directional spectrum is simply an integral of directional spectrum over all directions:

$$
\int_{0}^{\infty}F(k)dk = \int_{0}^{2\pi}\int_{0}^{\infty}F(k,\theta)dkd\theta
$$

Be careful about units here. All spectrum integrals must come up at $m^2$. Thus, $F(k)$ has units of $m^3$ and $F(k,\theta)$ has units of $m^3$ $rad^{-1}$.

If you are considering polar (spectral bins scaling with $k\theta$ instead of $\theta$) directional wavenumber spectrum such that:

$$
\int_{0}^{2\pi}\int_{0}^{\infty}F(k,\theta)k\ dk\ d\theta = \sigma^2
$$

then $F(k,\theta)$ has units of $m^4$ $rad^{-1}$.