What date is that again?
Pyth, 118 bytes

M++28@j15973358 4G&amp;qG2!%H4FN.pmv&gt;dqhd\0cz\.I&amp;&amp;&amp;hN&lt;hN13eN&lt;eNhgFPNaYK+++*365JhtN/+3J4smghdJthNeNInK60aY-K+12&gt;K60;-eSYhSY


Try it online: Demonstration or Test Suite.

Necessary knowledge of Julian and Gregorian Calendars

Julian and Gregorian Calendar are quite similar. Each calendar divides a year into 12 months, each containing of 28-31 days. The exact days in a month are [31, 28/29 (depends on leap year), 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]. The only difference between the calendars are their definition of a leap year. In the Julian Calendar any year divisible by 4 is a leap year. The Gregorian Calendar is a bit more specific. Any year divisible by 4 is a leap year, except the year divisible by 100 and not divisible by 400. 

So in the 20th century only one year is different. The year 1900, which is a leap year in the Julian Calendar, but not a leap year in the Gregorian Calendar. So the only date, that exists in the one calendar but not in the other calendar is the day 29.02.1900. 

Because of the different leap year definition, there's a difference between a date in the Julian Calendar and the Gregorian Calendar. 12 days difference for a date before the 29.02.1900, and 13 days difference for dates after the 29.02.1900. 

Simplified Pseudo-Code

Y = []  # empty list
for each permutation N of the input date:
   if N is valid in the Julian Calendar:
      K = number of days since 0.01.1900
      append K to Y
      if K != 60:  # 60 would be the 29.02.1900
         L = K - (12 if K &lt; 60 else 13) 
         append L to Y
print the difference between the largest and smallest value in Y


Detailed Code Explanation

The first part M++28@j15973358 4G&amp;qG2!%H4 defines a function g(G,H), which calculates the number of days in month G of a year H in the Julian Calendar. 

M                            def g(G,H): return
      j15973358 4               convert 15973358 into base 4
     @           G              take the Gth element
  +28                           + 28
 +                &amp;qG2!%H4      + (G == 2 and not H % 4)


And the next part is just the for loop, and the ifs. Notice that I interpret N in the format (month, year, day). Just because it saves some bytes. 

FN.pmv&gt;dqhd\0cz\.
             cz\.        split input by "."
    mv&gt;dqhd\0            map each d of ^ to: eval(d[d[0]=="0":])
FN.p                     for N in permutations(^):

I&amp;&amp;&amp;hN&lt;hN13eN&lt;eNhgFPN   
I                          if 
    hN                        month != 0
   &amp;                          and
      &lt;hN13                   month &lt; 13
  &amp;                           and
           eN                 day != 0
 &amp;                            and
             &lt;eNhgFPN         day &lt; 1 + g(month,year):

aYK+++*365JhtN/+3J4smghdJthNeN
          JhtN                    J = year
     +*365J   /+3J4               J*365 + (3 + J)/4
    +              smghdJthN      + sum(g(1+d,year) for d in [0, 1, ... month-2])
   +                        eN    + day
  K                               K = ^
aYK                               append K to Y

InK60aY-K+12&gt;K60            
InK60                             if K != 60:
     aY-K+12&gt;K60                    append K - (12 + (K &gt; 60)) to Y

;-eSYhSY
;          end for loop
 -eSYhSY   print end(sorted(Y)) - head(sorted(Y))