2. Houses on the left side of the street have odd, and houses on the right side of the street have even house numbers. The sum of all house numbers on one side of the street is 1369, and on the other side 2162. How many houses are there on that street?
First method:

On one side of the street, the house numbers are odd $(1,3,5, \ldots)$, and on the other side, they are even $(2,4,6, \ldots)$. Since the sum of even numbers is always even, it is clear that the sum of the odd house numbers is 1369, and the sum of the even house numbers is 2162.

Let $n$ be the number of houses on the odd side. The first house number on this side is 1, and the last is $2 n-1$.

Using the Gaussian principle, we get $1+3+5+\ldots+(2 n-1)=(1+2 n-1) \cdot n: 2=1369$, or $n \cdot n=1369$. The desired number must be between 30 and 40 because $30 \cdot 30=900$, and $40 \cdot 40=1600$. Since the last digit of this product is 9, the unit digit of the number $n$ must be either 3 or 7.

By checking, we find that $n=37$ (since $33 \cdot 33=1089$).

Let $m$ be the number of houses on the even side. The first house number on this side is 2, and the last is $2 m$. Using the Gaussian principle, we get $2+4+6+\ldots+2 m=(2+2 m) \cdot m: 2=2162$, or $(m+1) \cdot m=2162$. By factoring 2162 into prime factors, we get $2162=2 \cdot 23 \cdot 47=47 \cdot 46$. From this, we conclude that $m=46$.

On one side of the street, there are 37 houses, and on the other side, there are 46 houses. In the street, there are 83 houses.