How can be current increased with decreasing voltage? Here are the voltage and current shown for the primary winding of a transformer or an ideal inductor when the voltage is sinusodial.
My question is why the current is increasing when the voltage is decreasing as shown in the black box ? Its obvious that current should be increasing with increasing voltage and current should be decreasing with decreasing voltage but here the current is increasing for decreasing voltage.
And the current decreases during the negative voltage cycles only.
What is the the reason behind all this ?

 A: 
Its obvious that current should be increasing with increasing voltage and current should be decreasing with decreasing voltage but here the current is increasing for decreasing voltage.

It is not obvious, because if current is changing in time (electrons accelerate), intensity of current is affected not only by voltage, but also by induced electric field of the accelerated charges (forming the changing current). The more inductance, the greater the effect.
The obvious thing would be that current is proportional to net sum of voltage drop and induced emf along the path in direction of the current. Let the current flow from 1 to 2, passing some coiled wire in between.
*--- (some coiled wire) ---*
1                          2
potential V_1              potential V_2

This is sometimes called generalized Ohm's law: let emf for trip from 1 to 2 be $EMF(1\to 2)$ (non-standard notation). Then the current $I$ obeys
$$
RI = V_1-V_2 + EMF(1\to 2).
$$
where $R$ is ohmic resistance of the path $1\to 2$.
EMF for coiled wire is proportional to $dI/dt$. If there is enough turns of the wire,  EMF is important and current can't be assumed to be proportional to voltage alone. It is proportional to voltage plus emf.
A: You haven't said what is attached to the secondary of this transformer. The plot you showed is more-or-less consistent with this being an open circuit (i.e. no load attached).
With no load on the secondary, the primary just acts as an inductor. And as you know, the defining equation for an inductor is
$$V = L\frac{dI}{dT}$$
or
$$\frac{dI}{dT}=\frac{V}{L}.$$
This means that whenever $V>0$, an increasing current is exactly what you should expect regardless of whether $\frac{dV}{dt}>0$ or $\frac{dV}{dt}<0$.
The exact shape of the curve doesn't look quite right for this scenario. We would expect a sinusoidal shape for the current waveform when the voltage waveform is sinusoidal. 
Or, if the core were saturating, we'd expect the current to increase more, rather than less, than in the linear (non-saturating) case.
The only think I can imagine is that there's some load on the secondary, or some device in series with the primary, that limits the current below the expected values for an ideal transformer.
Or the drawing was just made free-hand and isn't expected to give the exact waveforms.
Using a SPICE simulator program, it's easy to get a view of what the curve should look like (this assumes a 1 V peak, 1 Hz sine wave source and 1 H inductor):

You can see the current waveform is symmetric about the point at 250 ms, rather than being stretched upwards as in the plot you were given.
