This is a very bad write-up. But, here we go... Part a) The construction for $n$ roadblocks is easy, just read the vertical lines from left to right, and alternatively close 2-nd and 3-rd crossroads on these vertical lines. We will now show that $n$ roadblocks are necessary. Regard the boulevards as the gridlines of a $3\times(n-1)$ grid. Color the grid as in the following figure (black and white alternating in the first and last row with leftmost squares receiving black; the middle row is all white) [figure is attached.... I don't know how to add images in the middle of the passage, lol] Notice that at least one corner of each black square is a closed crossroad, or otherwise we have a cycle. Therefore, if $n$ is even, we are immediately done. If $n$ is odd, the above argument only shows that we need at least $n-1$ roadblocks. Suppose that it is possible to close exactly $n-1$ crossroads. Read the first row from left to right, and pair up the $k$-th black square with $k$-th white square. By the above argument, there is exactly one roadblock on each black square on the first row, and each white square on the first row. It follows there is a roadblock on one of the two vertices shared by the $k$-th black square and the $k$-th white square. The same argument holds for the third row. Now, look at the 3 leftmost vertical roads. There is no roadblock on the first and third leftmost vertical roads, and exactly two roadblocks on the second. Hence, there will be a cycle, contradiction. (I'm pretty sure this part has a much much better solution, and I would love to see one )) ) $\color{black} \rule{14cm}{1pt}$ Part b) This part is fairly straightforward. Consider a graph $G$ whose vertices are crossroads with an edge between two crossroads if and only if there is exactly one street joining them (hence, $G$ is a grid graph). Now, remove those vertices that correspond to closed crossroads, and let $H$ be the resulting graph. If the conditions in the problem statement are satisfied, then $H$ should be a tree. Since, $G$ has $4n$ vertices and $n$ of them are removed, $H$ has exactly $3n-1$ edges. Since $G$ has $3n+4(n-1)=7n-4$ edges, that means $4n-3$ edges are removed. However, notice that no two removed vertices are adjacent or otherwise there will be a "trapped" street. Also, corner vertices are not removed. Therefore, whenever a vertex is removed from $G$, number of edges either decrease by $4$ or $3$. Since the end result is that $4n-3$ edges are removed along with $n$ vertices, this means that exactly three of the vertices on the border are removed.

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Nice problem. The equality case for a) is the same as the above post. I will post a bit different sol of the bound of a) and b). a) We will do induction on $n$. So note that if the leftmost vertical boulevard has at least one deleted point, use the hypothesis for $n-1$. Otherwise the adjacent boulevard to the leftmost boulevard has at least two deleted points (or we have a "unit" rectangle without deleted vertices, so this is a cycle). Now use the hypothesis for $n-2$, and we are done. b) Consider the grid graph, formed by the crossroads as vertices and streets as edges. In the beginning we have $4n$ vertices and $4(n-1)+n.3=7n-4$ edges. Let the number of deleted border vertices be $y$ and the number of internal deleted vertices be $x$. When we delete an internal vertex, we delete 4 edges, and when we delete border vertex, we delete 3 edges. At the end we have connected graph with $3n$ vertices without cycles, and thus this is a tree, so we have $3n-1$ edges. So $4n-3$ edges have been deleted. Then $4x+3y=4n-3$, and using a) that $x+y=n$, so $x=n-3$ and $y=3$, which is what we wanted.

Consider a graph where the $7n - 4$ streets are vertices and there exists a path between two streets in the graph iff there exists one in the city. Instead of constructing a complete graph among the streets meeting at each crossroads, we will construct a path graph, so that connectedness still behaves the same way but there are no "fake cycles". Notice that each corner crossroad corresponds to $1$ edge, each noncorner border crossroad to $2$ edges, and each middle crossroad to $3$ edges. In particular, the graph initially has $10n - 8$ edges. After closure of crossroads, this graph should be a forest. a) For the bound, assume FTSOC that at most $n-1$ crossroads are closed. Since the border is a cycle, at least one border crossroad must be closed. In particular, the graph will still have at least $10n - 8 - 3(n-2) - 2 = 7n - 4$ edges and hence must contain a cycle. For the construction, close the $2$nd crossroad from the top in odd-numbered columns and the $3$rd crossroads from the top in even-numbered columns (where columns are $3$ streets long and rows are $n-1$ streets long). b) By the additional condition the graph in question must be a tree, which immediately tells us that we must close $n-3$ middle crossroads and $3$ noncorner border crossroads to be left with $7n-5$ edges.