Fill missing rows in a python pandas dataframe with repetitive pattern

You need create groups some way - here is used difference of values # and comparing with >1 by Series.le, then is used GroupBy.apply with Series.reindex:

df1 = (df.groupby(df['#'].diff().lt(1).cumsum())
         .apply(lambda x: x.set_index('#').reindex(range(1, 5)))
         .reset_index(level=0, drop=True)
         .reset_index())

print (df1)
    #  foo  bar
0   1  1.2  3.4
1   2  4.5  6.7
2   3  1.3  2.5
3   4  5.6  7.3
4   1  3.4  5.8
5   2  5.7  8.9
6   3  NaN  NaN
7   4  2.4  2.6
8   1  6.7  8.4
9   2  NaN  NaN
10  3  6.9  4.2
11  4  4.2  1.2

Another idea is create MultiIndex and reshape by unstack and stack:

df = (df.set_index(['#', df['#'].diff().lt(1).cumsum()])
       .unstack()
       .reindex(np.arange(4)+1)
       .stack(dropna=False)
       .sort_index(level=1)
       .reset_index(level=1, drop=True)
       .reset_index())
print (df)
    #  foo  bar
0   1  1.2  3.4
1   2  4.5  6.7
2   3  1.3  2.5
3   4  5.6  7.3
4   1  3.4  5.8
5   2  5.7  8.9
6   3  NaN  NaN
7   4  2.4  2.6
8   1  6.7  8.4
9   2  NaN  NaN
10  3  6.9  4.2
11  4  4.2  1.2

We can mark each group of 1,2,3,4 with eq and cumsum.

Then we groupby on these groups and use reindex and finally concat them back together.

s = df['#'].eq(4).shift().cumsum().bfill()

pd.concat(
    [d.set_index('#').reindex(np.arange(4)+1) for _, d in df.groupby(s)]
).reset_index()

Output

    #  foo  bar
0   1  1.2  3.4
1   2  4.5  6.7
2   3  1.3  2.5
3   4  5.6  7.3
4   1  3.4  5.8
5   2  5.7  8.9
6   3  NaN  NaN
7   4  2.4  2.6
8   1  6.7  8.4
9   2  NaN  NaN
10  3  6.9  4.2
11  4  4.2  1.2

---

Note: if you would have a 4 as missing value in your # column, this method would fail.