I think this is possible. So to accommodate/test your proposed outcome we would simply NOT place the string over the interim poles (the poles other than the first, last, and center) and the string would drop below them - correct?

Isn't this all just a question of scale though? Do you really think there is no circular curved surface which would cause a gap between the taught string and the interim poles?

No, it isn't a question of scale.

The problem is that the string would sag, with that sag being more than the drop due to curvature.

The other important point is that it would not cause a gap between the string and poles. Instead the top of the poles would be above the string.

In order to compensate for the sag, you need to know the material properties of the string, primarily its linear density, and want that to be as constant as possible.

You then need to know exactly what the tension is.

Ditching all bar the end and middle poles, you have a 100 m span for the string to sag in and for the curvature to cause a drop.

The drop due to curvature over such a distance is relatively easy to calculate. As an approximation, h=d^2/(2*R).

So for a distance of 100 m (from the middle pole to the edge) you would expect a drop of ~0.8 mm.

This is tiny.

Conversely, the string would follow a catenary curve. This can be described as y=a*(cosh(x/a)-1).

Where a=T/(lambda*g), where T is the tension in the string and lambda is the linear density.

We can further simply this by noting lambda = rho*A, where A is the cross sectional area, and we can convert T into a stress (pressure) by noting T=sigma*A.

This means we can simply as a=(sigma*A)/(rho*A*g) = sigma/(rho*g).

And the key property here would be the tensile strength divided by density. As the tensile strength is the maximum value of sigma you can use without breaking the string.

This is know as the specific strength.

For nylon, the value is roughly 70 kN m / kg, giving a drop of 0.7 m over this distance.

But you wouldn't actually want it at breaking point. If we leave plenty of margin and only have the string at 50% of its breaking point, it would give a drop of 1.4 m.

This is important due to the control of the tension.

If that tension is off by just 0.1%, then the drop will change by 1.4 mm.

That is almost 2 times the drop you are trying to measure.

Switching to a stainless steel wire isn't any better due to its much larger density.

To have any hope you would need something like zylon.

That has a specific strength of ~3800. That would give a drop (assuming loaded to 50% of breaking stress) of 25 mm. Which is still over 30 times the drop due to curvature.

Also, with this, if you had a string that was just 1 mm in diameter, this would require a force equivalent to ~ 800 kg. And you would need to make sure that massive force does not cause any deviation of the pole. That means it cannot compress the pole, nor can it bend it, bury into the ground, or put it out of perfect vertical alignment.

Which will be quite difficult to do considering you are trying to find a drop of 0.8 mm.

It is simply not pratical.