Beginner's attempt for Project Euler problem 12 (highly divisible triangle number)
Iteration, flow control, naming, and clarity

Your iteration is a bit clumsy.

It seems that target is an arbitrarily chosen number that you picked just to act as a sufficiently large second parameter to xrange().  If you want an unbounded counting loop, itertools.count(1) would be better.

Instead of else: counter = 0, you should just set counter = 0 in the right place — just before the inner loop.

Avoid using sys.exit() — it's like killing your program with a sledgehammer.  You should structure your code so that it terminates naturally.

from itertools import count

for value in count(1):
    x = (value * (value + 1))/2
    for divisor in xrange(1, x+1):
        product = x % divisor
        if product == 0:
            counter += 1

    if counter &gt;= 500:
        break

print x


There is an idiom for counting items that meet some criterion: sum() with a generator expression.

from itertools import count

for value in count(1):
    x = (value * (value + 1)) / 2
    if 500 &lt;= sum(1 for divisor in xrange(1, x + 1) if x % divisor == 0):
        print x
        break


As you noted yourself, your variables are poorly named.  product, in particular, should be called remainder.  counter is vague; divisor_count would be more helpful.

What you want to express is a loop over the triangle numbers.  For even greater clarity, I'd break that out into a triangle number generator function.

from itertools import count

def triangle_numbers():
    """Generator of triangle numbers, starting with 1, 1+2, 1+2+3, ..."""
    n = 0
    for i in count(1):
        n += i
        yield n

def divisor_count(n):
    """Count of the divisors of n"""
    return sum(1 for divisor in xrange(1, n + 1) if n % divisor == 0)

print next(t for t in triangle_numbers() if divisor_count(t) &gt;= 500)


Mathematics

Note that \$\dfrac{n(n+1)}{2}\$ is a product of two coprimes.  As explained here, the divisors of such a product can be computed based on the divisors of each of its two known factors.

from itertools import count

def divisor_count(n):
    """Count of the divisors of n"""
    return sum(1 for divisor in xrange(1, n + 1) if n % divisor == 0)

for n in count(1):
    tn = n * (n + 1) // 2
    # n and (n + 1) are relatively prime.  Therefore, if n is even,
    # the factors of tn can be derived from the factors of n/2 and
    # the factors of n+1. If n is odd, then the factors of tn can be
    # derived from the factors of n and the factors of ((n+1)/2).
    tn_divisor_count = (
        divisor_count(n // 2) * divisor_count(n + 1) if n % 2 == 0 else
        divisor_count(n) * divisor_count((n + 1) // 2)
    )
    if tn_divisor_count &gt;= 500:
        print tn
        break