SOLUTION: On a certain street, the house numbers incrementally increase on one side and come back down the other side. So, on one side of the street the house numbers are 1,2,3 etc. House nu

The answer is probably supposed to be 93 houses. I'll show you
how I got that below. However first I must say that there
can be no solution if there must be the same number of 
houses on each side of the street. That's because every even 
numbered house would be opposite an odd numbered house, and 
the problem states that two even numbered houses, 14 and 80,
are opposite each other.  That cannot be if there are the same
number of houses on each side of the street  

Why? Take an example of only 10 houses, with 5 on each side:

 1  2  3  4  5 
10  9  8  7  6

Whatever the last number on the first side is, then the house 
across the street must be numbered 1 more than it.  If the last
number on the first side is ODD, as in this case, then the house 
across the street from it must have an EVEN number, since it must 
be one more.  And as you see each even numbered house is across
from an odd numbered house.

1 2 3 4 
8 7 6 5

if the the last number on the first side is even, as in this case, 
then the house across the street must be odd, for it is one more.

The only way this problem could have a solution would be for one
side of the street to have at least one more house than the other side.
That is, one or more of the houses would have to have a vacant lot 
across the street from it.  

If even numbered houses are across the street from even numbered
houses it would have to be something like this:

1 2 3 4 5
9 8 7 6 V

where the last house, number 5, is opposite V, a vacant lot. 
But if we allow 1 vacant lot, then why not allow 2 or 3 or
more vacant lots as well?  Then there could be many different
answers, for instance:

1 2 3 V 4
7 V V 6 5

Whoever made up this problem didn't think of that, and mistakenly 
put two even numbered houses across from each other, when they
should have had an odd number across from an even number, avoiding
the need for any house across from a vacant lot.  But, OK, let's assume 
there is 1 vacant lot across from one of the houses right of the
14th.  Then here is the situation with the 14 houses on the far left 
side of the street:

 1  2  3  4  5  6  7  8  9 10 11 12 13 14 ...
93 92 91 90 89 88 87 86 85 84 83 82 81 80 ...
  
So the answer would be 93 if we assume just 1 vacant lot which is to 
the right of the leftmost 14.  This is probably what the author of
the problem had in mind.  But we could assume there were 2 vacant 
lots and the 2nd vacant lot was left of the 14th house and then there 
would only be 92 houses, or 3 vacant lots and only 91 house, etc., etc.
There is no doubt that this problem is botched!  Mention this to your
teacher that there cannot be the same number of houses on each side
of the street without assuming one or more house with a vacant lot 
opposite it. 

Edwin