Are we slightly lighter during the day and slightly heavier at night, owing to the force of the Sun's gravity?

This diagram shows the Earth rotating round the Sun at it's orbital velocity $v$. That is the centre of the Earth is orbiting around the Sun at velocity $v$. NB the scale is rather fanciful - don't take it literally! I'll also assume the orbit is circular, and for convenience I'll ignore the Earth's rotation i.e. assume it's tidally locked.

To calculate the orbital velocity at the centre of the Earth, $v$, we just note that the centripetal acceleration must be the same as the gravitational acceleration of the Sun so:

$$ \frac{v^2}{r} = \frac{GM}{r^2} $$

which gives:

$$ v^2 = \frac{GM}{r} \tag{1} $$

which is a well known result. Now consider the point on the Earth's surface nearest the Sun i.e. the black dot. The acceleration due to Earth's gravity is the usual $9.81 m/s^2$, but there will be a correction due to the fact the point is $r_e$ metres nearer the Sun. Let's calculate that correction.

The gravitational acceleration due to the Sun at the black dot is:

$$ a_g = \frac{GM}{(r - r_e)^2} $$

The centripetal acceleration due to the motion of the point around the Sun is:

$$ a_c = \frac{v^2}{r - r_e} $$

where because I've assumed the Earth is tidally locked the velocity $v$ is just the Earth's orbital velocity given by equation (1). If we substitute for this we get:

$$ a_c = \frac{GM}{r(r - r_e)} $$

So the correction to the acceleration at the black dot is:

$$\begin{align} \Delta a &= a_g - a_c \\ &= \frac{GM}{(r - r_e)^2} - \frac{GM}{r(r - r_e)} \\ &= GM \left( \frac{r_e}{r(r - r_e)^2} \right) \\ &\approx GM \frac{r_e}{r^3} \end{align}$$

where the last approximation is because $r \gg r_e$ so $r - r_e \approx r$. Putting in the numbers we get:

$$ \Delta a \approx 2.5 \times 10^{-7} m/s^2 $$

So the fractional change in the weight of an object due to the Sun is:

$$ \frac{2.5 \times 10^{-7}}{g} \approx 2.6 \times 10^{-8} $$

and the object is 0.0000026% lighter. Interestingly if you go through the working for the far side of the Earth you get exactly the same result i.e. the object on the far side is also 0.0000026% lighter. In fact this is why the tidal forces of the Sun (and Moon of course) raise a bulge on both the near and far sides of the Earth.

Incidentally, I note that Christoph guesstimated a correction of $10^{-7}$ and he was pretty close :-)

Yes your weight will change. The moon will have a bigger impact than the sun, so you need to look at the position of the moon to decide when you will be heaviest (basically - you are lighter when the moon is overhead, or on the opposite side of the earth; and heaviest when it is on the horizon. So a full moon rising makes you fat...)

The effect (the variation in $g$ over the course of a day) has been measured very carefully:

This figure is on page 93 of "Practical Physics" by Gordon Squires (a classical book, and one that I highly recommend). The method used is a beautiful example of careful experimental work, where the speed with which a corner cube drops is measured using an interferometric measurement. The active vibration damping of the reference mirror, clock calibration, etc are a pleasure to read - especially when you think this was done over 30 years ago. They quote a residual error of $60 nm/s^2$ or about 6 ppb. That is stunning.

Note - there is a clear asymmetry here: it is as though the tides don't pull evenly. I believe the reason for this is the relative tilt between the earth's axis and the plane of rotation of the sun and moon. I explained this with a diagram in my answer to another question.

Let's simplify.

Let's eliminate the Moon.

Let's get rid of the Sun temporarily.

Let's replace the Earth with an equivalent mass-and-density perfect sphere of iron that is neither moving linearly nor spinning or revolving in any way.

We place two 1KG iron test masses on opposite sides of the Iron Earth, suspended 1 M above the surface by identical springs. Each experiences a 9.8 N force towards the center of Iron Earth. The amount of distortion of each spring is identical.

Fine.

Now we add in an Iron Sun. Let's connect the Iron Earth and the Iron Sun with a perfectly rigid, massless bar that prevents relative motion between them. Again, no spinning, etc.

Now what are the forces on the test masses? Call the one close to Iron Sun the noon mass.

The noon mass has a force of 9.8 N towards the center of the Iron Earth, and an opposing force of gravity from the Iron Sun towards the Iron Sun.

The midnight mass has a force of 9.8 N towards the center of the Iron Earth and a slightly smaller than before because it is farther from the Iron Sun force of gravity towards the Iron Sun, which adds to the force towards the Iron Earth.

So in this scenario, the springs are stretched by different amounts. The noon spring is stretched less than in our first experiment, and the midnight spring is stretched more than in our first experiment. There is a slight difference in the magnitude of the differences in stretchiness because of the diameter of the Earth.

Now let's eliminate the impossible massless bar connecting Iron Earth and Iron Sun, and replace it with two rockets, one on each planet, that magically pushes the Iron Earth away from the Iron Sun and vice versa exactly enough to counteract the force of gravity from each on the other. So again, they are stationary with respect to each other.

How do the springs change?

They don't. This is the same situation as before. The compression force opposing gravity that previously held Iron Earth and Iron Sun apart despite a huge force of gravity between them has been replaced by a propulsion force; this is a difference that makes no difference.

Now we turn off the rockets, so Iron Earth and Iron Sun start falling straight towards each other. What happens to the springs in the first minute?

This is the question you actually need to answer. If you work it out, you'll see that the accelerating movement of the Iron Earth towards the Iron Sun is exactly enough to compress the noon spring and stretch the midnight spring. There will still be a small difference, but it will be to the difference in the force of gravity from the Sun across the diameter of the Earth; hence my confusion in my original comment to your question. That's the difference I thought you were asking about.

The Earth is of course not falling in a straight line towards the Sun, but that's irrelevant; the acceleration vector is in that direction and that's what matters.

Fun bonus question: If we turn the rockets off at exactly the same time (as observed by an observer at rest with respect to the planets, halfway between them) does the Iron Earth start falling towards the Iron Sun immediately, or do we have to wait eight minutes for the gravity to get from the Iron Sun to the Iron Earth at the speed of light? If it happens instantly, is this a way of communicating faster than the speed of light? Isn't that supposed to be impossible? See if you can work out what happens and why.