Let $T_{1}, T_{2}, \ldots, T_{n}$ be the towns enumerated from left to right. Observe first that, if town $T_{i}$ can sweep away town $T_{j}$, then $T_{i}$ also can sweep away every town located between $T_{i}$ and $T_{j}$.
We prove the problem statement by strong induction on $n$. The base case $n=1$ is trivial.
For the induction step, we first observe that the left bulldozer in $T_{1}$ and the right bulldozer in $T_{n}$ are completely useless, so we may forget them forever. Among the other $2n-2$ bulldozers, we choose the largest one. Without loss of generality, it is the right bulldozer of some town $T_{k}$ with $k<n$.
Surely, with this large bulldozer $T_{k}$ can sweep away all the towns to the right of it. Moreover, none of these towns can sweep $T_{k}$ away; so they also cannot sweep away any town to the left of $T_{k}$. Thus, if we remove the towns $T_{k+1}, T_{k+2}, \ldots, T_{n}$, none of the remaining towns would change its status of being (un)sweepable away by the others.
Applying the induction hypothesis to the remaining towns, we find a unique town among $T_{1}, T_{2}, \ldots, T_{k}$ which cannot be swept away. By the above reasons, it is also the unique such town in the initial situation. Thus the induction step is established.

We start with the same enumeration and the same observation as in Solution 1. We also denote by $\ell_{i}$ and $r_{i}$ the sizes of the left and the right bulldozers belonging to $T_{i}$, respectively. One may easily see that no two towns $T_{i}$ and $T_{j}$ with $i<j$ can sweep each other away, for this would yield $r_{i}>\ell_{j}>r_{i}$.
Clearly, there is no town which can sweep $T_{n}$ away from the right. Then we may choose the leftmost town $T_{k}$ which cannot be swept away from the right. One can observe now that no town $T_{i}$ with $i>k$ may sweep away some town $T_{j}$ with $j<k$, for otherwise $T_{i}$ would be able to sweep $T_{k}$ away as well.
Now we prove two claims, showing together that $T_{k}$ is the unique town which cannot be swept away, and thus establishing the problem statement.

Claim 1. $T_{k}$ also cannot be swept away from the left.
Proof. Let $T_{m}$ be some town to the left of $T_{k}$. By the choice of $T_{k}$, town $T_{m}$ can be swept away from the right by some town $T_{p}$ with $p>m$. As we have already observed, $p$ cannot be greater than $k$. On the other hand, $T_{m}$ cannot sweep $T_{p}$ away, so a fortiori it cannot sweep $T_{k}$ away.

Claim 2. Any town $T_{m}$ with $m \neq k$ can be swept away by some other town.
Proof. If $m<k$, then $T_{m}$ can be swept away from the right due to the choice of $T_{k}$. In the remaining case we have $m>k$.
Let $T_{p}$ be a town among $T_{k}, T_{k+1}, \ldots, T_{m-1}$ having the largest right bulldozer. We claim that $T_{p}$ can sweep $T_{m}$ away. If this is not the case, then $r_{p}<\ell_{q}$ for some $q$ with $p<q \leqslant m$. But this means that $\ell_{q}$ is greater than all the numbers $r_{i}$ with $k \leqslant i \leqslant m-1$, so $T_{q}$ can sweep $T_{k}$ away. This contradicts the choice of $T_{k}$.

We separately prove that (i) there exists a town which cannot be swept away, and that (ii) there is at most one such town. We also make use of the two observations from the previous solutions.
To prove $(i)$, assume contrariwise that every town can be swept away. Let $t_{1}$ be the leftmost town; next, for every $k=1,2, \ldots$ we inductively choose $t_{k+1}$ to be some town which can sweep $t_{k}$ away. Now we claim that for every $k=1,2, \ldots$, the town $t_{k+1}$ is to the right of $t_{k}$; this leads to the contradiction, since the number of towns is finite.
Induction on $k$. The base case $k=1$ is clear due to the choice of $t_{1}$. Assume now that for all $j$ with $1 \leqslant j<k$, the town $t_{j+1}$ is to the right of $t_{j}$. Suppose that $t_{k+1}$ is situated to the left of $t_{k}$; then it lies between $t_{j}$ and $t_{j+1}$ (possibly coinciding with $t_{j}$ ) for some $j<k$. Therefore, $t_{k+1}$ can be swept away by $t_{j+1}$, which shows that it cannot sweep $t_{j+1}$ away - so $t_{k+1}$ also cannot sweep $t_{k}$ away. This contradiction proves the induction step.
To prove (ii), we also argue indirectly and choose two towns $A$ and $B$ neither of which can be swept away, with $A$ being to the left of $B$. Consider the largest bulldozer $b$ between them (taking into consideration the right bulldozer of $A$ and the left bulldozer of $B$ ). Without loss of generality, $b$ is a left bulldozer; then it is situated in some town to the right of $A$, and this town may sweep $A$ away since nothing prevents it from doing that. A contradiction.