Original wire resistance

The total resistance of the wire is $R = 20\,\Omega.$

Wire divided into 10 equal parts

Since resistance is directly proportional to length, each of the 10 equal parts has resistance $R_0 = \frac{20}{10} = 2\,\Omega.$

Make pairs and connect each pair in parallel

Two equal resistors of $2\,\Omega$ each are connected in parallel.

Their equivalent resistance is $R_p = \frac{2\times 2}{2+2} = \frac{4}{4} = 1\,\Omega.$

How many such parallel pairs are formed?

From 10 parts, taking 2 in each pair gives $\frac{10}{2} = 5\text{ pairs}.$

So we get 5 parallel combinations, each of resistance $1\,\Omega$.

These resulting pairs are connected in series

In series, resistances add directly: $R_{eq} = 1+1+1+1+1 = 5\,\Omega.$

Final answer

$\boxed{5\,\Omega}$