I feel the reply is:

Five miles north of the equator.

Reasoning:

This is a riff on the classical North Pole riddle, however the twist is that you just appear to make a full circuit, so it’s tempting to say “anyplace”. However, as a result of the earth is (roughly) a sphere, 10 miles east and 10 miles west are the identical distance if and provided that they’re walked on the similar latitude, or equal latitudes within the Northern/Southern hemispheres. Since the east and west segments are walked 10 miles north/south from one another, the one manner this will occur is that if these two segments are on reverse sides of the equator. Hence you will need to begin 5 miles north of the equator.

Update: Based on @Bass’s reply, I used to be curious concerning the precise calculation. Turns out it is actually not that arduous.

First, let’s approximate earth with a sphere of radius 4,000 miles…the quantity is a bit excessive, however that is all an approximation anyway. Let $theta$ be the angle between the axis of the sphere by the south pole, and the axis of the sphere by your place to begin, as on this diagram (flagrantly to not scale):

First be aware that the arc size between the beginning and finish of the stroll on this circle is 10 miles, so we all know that the angle between the beginning and finish of the stroll is $theta – phi = frac{10}{4000} = frac{1}{400}$ radians.

We can now simply decide the angle of longitude our 10 mile walks take. At the start line, we’re strolling round a circle of radius $R = 4000 sin theta$ for 10 miles, so the angle of longitude is $frac{10}{4000 sin theta}$. Note the angle equals arclength over radius formulation actually helps us right here, because it takes care of a number of passes across the circle with none effort on our half. Similarly, the angle of longitude on the backside stroll is $frac{10}{4000 sin (theta – frac{1}{400})}$.

So for this stroll to be doable, we will need to have $$ frac{10}{4000 sin (theta – frac{1}{400})} – frac{10}{4000 sin theta}$$ be a constructive integer a number of of $2pi$. Since these are each steady features of $theta$ away from $frac{1}{400}$, it’s straightforward to see by the intermediate worth theorem (h/t @JaapScherphuis) that this perform takes on all above its minimal.