Rounding is typically done on floating point numbers, and here there are three basic functions you should know: round (rounds to the nearest integer), math.floor (always rounds down), and math.ceil (always rounds up).

You ask about integers and rounding up to hundreds, but we can still use math.ceil as long as your numbers smaller than 2⁵³. To use math.ceil, we just divide by 100 first, round up, and multiply with 100 afterwards:

Copy>>> import math
>>> def roundup(x):
...     return int(math.ceil(x / 100.0)) * 100
... 
>>> roundup(100)
100
>>> roundup(101)
200

Dividing by 100 first and multiply with 100 afterwards "shifts" two decimal places to the right and left so that math.ceil works on the hundreds. You could use 10**n instead of 100 if you want to round to tens (n = 1), thousands (n = 3), etc.

An alternative way to do this is to avoid floating point numbers (they have limited precision) and instead use integers only. Integers have arbitrary precision in Python, so this lets you round numbers of any size. The rule for rounding is simple: find the remainder after division with 100, and add 100 minus this remainder if it's non-zero:

Copy>>> def roundup(x):
...     return x if x % 100 == 0 else x + 100 - x % 100

This works for numbers of any size:

Copy>>> roundup(100)
100
>>> roundup(130)
200
>>> roundup(1234567891234567891)
1234567891234567900L

I did a mini-benchmark of the two solutions:

Copy$ python -m timeit -s 'import math' -s 'x = 130' 'int(math.ceil(x/100.0)) * 100'
1000000 loops, best of 3: 0.364 usec per loop
$ python -m timeit -s 'x = 130' 'x if x % 100 == 0 else x + 100 - x % 100'
10000000 loops, best of 3: 0.162 usec per loop

The pure integer solution is faster by a factor of two compared to the math.ceil solution.

Thomas proposed an integer based solution that is identical to the one I have above, except that it uses a trick by multiplying Boolean values. It is interesting to see that there is no speed advantage of writing the code this way:

Copy$ python -m timeit -s 'x = 130' 'x + 100*(x%100>0) - x%100'
10000000 loops, best of 3: 0.167 usec per loop

As a final remark, let me also note, that if you had wanted to round 101–149 to 100 and round 150–199 to 200, e.g., round to the nearest hundred, then the built-in round function can do that for you:

Copy>>> int(round(130, -2))
100
>>> int(round(170, -2))
200

You need divide by 100, convert to int and last multiple by 100:

df['new_values'] = (df['old_values'] / 100).astype(int) *100

Same as:

df['new_values'] = (df['old_values'] / 100).apply(np.floor).astype(int) *100

Sample:

df = pd.DataFrame({'old_values':[8450, 8470, 343, 573, 34543, 23999]})
df['new_values'] = (df['old_values'] / 100).astype(int) *100
print (df)
   old_values  new_values
0        8450        8400
1        8470        8400
2         343         300
3         573         500
4       34543       34500
5       23999       23900

EDIT:

df = pd.DataFrame({'old_values':[3, 6, 89, 573, 34, 23]})
#show output of first divide for verifying output
df['new_values1'] = (10000/df['old_values'])
df['new_values'] = (10000/df['old_values']).div(100).astype(int).mul(100)
print (df)
   old_values  new_values1  new_values
0           3  3333.333333        3300
1           6  1666.666667        1600
2          89   112.359551         100
3         573    17.452007           0
4          34   294.117647         200
5          23   434.782609         400

int(round(123,-2))

The int function can be used to convert a string or number to a plain integer.

def round_down(num, divisor):
    return num - (num%divisor)

In [2]: round_down(19,10)
Out[2]: 10

In [3]: round_down(19,5)
Out[3]: 15

In [4]: round_down(10,10)
Out[4]: 10

A general purpose solution, this allows rounding to an arbitrary resolution (well, other than zero of course, but a resolution of zero makes little sense ^(a)). For your case, you just need to provide 0.02 as the resolution, though other values are possible, as shown in the test cases.

# This is the function you want.

def roundPartial (value, resolution):
    return round (value / resolution) * resolution

# All these are just test cases, the first two being your own test data.

print "Rounding to fiftieths"
print roundPartial (0.405, 0.02)
print roundPartial (0.412, 0.02)

print "Rounding to quarters"
print roundPartial (1.38, 0.25)
print roundPartial (1.12, 0.25)
print roundPartial (9.24, 0.25)
print roundPartial (7.76, 0.25)

print "Rounding to hundreds"
print roundPartial (987654321, 100)

This outputs:

Rounding to fiftieths
0.4
0.42
Rounding to quarters
1.5
1.0
9.25
7.75
Rounding to hundreds
987654300.0

^(a) If you have the particular personality disorder that requires you to handle this possibility, just be aware that you're after the closest number that is a multiple of your desired resolution. Since the closest number to N (for any N) that is a multiple of 0 is always 0, you could modify the function as follows:

def roundPartial (value, resolution):
    if resolution == 0:
        return 0
    return round (value / resolution) * resolution

Alternatively, you could simply promise yourself not to pass zero as a resolution :-)