City of Mar del Plata is a square shaped $WSEN$ land with $2(n + 1)$ streets that divides it into $n \times n$ blocks, where $n$ is an even number (the leading streets form the perimeter of the square). Each block has a dimension of $100 \times 100$ meters. All streets in Mar del Plata are one-way. The streets which are parallel and adjacent to each other are directed in opposite direction. Street $WS$ is driven in the direction from $W$ to $S$ and the street $WN$ travels from $W$ to $N$. A street cleaning car starts from point $W$. The driver wants to go to the point $E$ and in doing so, he must cross as much as possible roads. What is the length of the longest route he can go, if any $100$-meter stretch cannot be crossed more than once? (The figure shows a plan of the city for $n=6$ and one of the possible - but not the longest - routes of the street cleaning car. See http://goo.gl/maps/JAzD too.)
[img]http://s14.postimg.org/avfg7ygb5/CPS_2012_P5.jpg[/img]
1. **Understanding the Problem:**
   - The city is divided into \( n \times n \) blocks by \( 2(n+1) \) streets.
   - Each block is \( 100 \times 100 \) meters.
   - Streets are one-way and adjacent streets have opposite directions.
   - The goal is to find the longest route from point \( W \) to point \( E \) without crossing any \( 100 \)-meter stretch more than once.

2. **Setting Up the Problem:**
   - Define \( a, b, c, d \) as the number of unit streets in the route oriented to \( SE, NW, SW, NE \) respectively.
   - Define \( A, B, C, D \) as the total number of unit streets in the city oriented to \( SE, NW, SW, NE \) respectively.

3. **Relations Between Variables:**
   - Since the car starts from \( W \) and must reach \( E \), we have:
     \[
     a = n + b \quad \text{and} \quad c = n + d
     \]
   - The total number of unit streets in each direction is:
     \[
     A = n + B \quad \text{and} \quad C = n + D
     \]

4. **Analyzing the Route:**
   - Consider the outdegree and indegree of the points:
     - For blue points (outdegree is 1 greater than indegree), there is one unit street leaving the point that can't be used.
     - For red points (indegree is 1 greater than outdegree), there is one unit street arriving at the point that can't be used.
   - There are \( 2n - 2 \) such colored points.

5. **Formulating the Inequality:**
   - The number of unused streets oriented to \( SE \) or \( SW \) is at least \( 2n - 2 \).
   - Therefore:
     \[
     a + c \leq A + C - (2n - 2)
     \]
   - Using the relations \( a = n + b \) and \( c = n + d \), we get:
     \[
     b + d \leq B + D - (2n - 2)
     \]

6. **Combining the Inequalities:**
   - Summing the inequalities:
     \[
     a + b + c + d \leq A + B + C + D - 4(n - 1)
     \]
   - This shows that at least \( 4(n - 1) \) unit streets remain uncrossed.

7. **Conclusion:**
   - The maximal route leaves \( 4(n - 1) \) unit streets uncrossed.

\(\blacksquare\)

The final answer is \( \boxed{ 4(n - 1) } \) unit streets remain uncrossed.