Given a non-negative integer Excel-style date code, return the corresponding "date" in any reasonable form that clearly shows year, month, and "day".

Trivial, you may think. Did you notice the "scare quotes"? I used those because Excel has some quirks. Excel counts days with number 1 for Januray 1st, 1900, but as if 1900 had a January 0th and a February 29th, so be very careful to try all test cases:

 Input → Output (example format)

     0 → 1900-01-00    Note: NOT 1899-12-31

     1 → 1900-01-01

     2 → 1900-01-02

    59 → 1900-02-28

    60 → 1900-02-29    Note: NOT 1900-03-01

    61 → 1900-03-01

   100 → 1900-04-09

  1000 → 1902-09-26

 10000 → 1927-05-18

100000 → 2173-10-14

Excel, 3(+7?)

=A1

with format

yyy/m/d

Pure port

Python 2, 116 bytes

from datetime import*

n=input()

print('1900-0'+'12--0209'[n>9::2],date(1900,1,1)+timedelta(days=n+~(n>60)))[0<n!=60]

Try it online!

k (kdb+ 3.5), 55 54 51 50 bytes

{$(`1900.01.00`1900.02.29,"d"$x-36526-x<60)0 60?x}

to test, paste this line in the q console:

k)-1@{$(`1900.01.00`1900.02.29,"d"$x-36526-x<60)0 60?x}'0 1 2 59 60 61 100 1000 10000 100000;

the output should be

1900.01.00

1900.01.01

1900.01.02

1900.02.28

1900.02.29

1900.03.01

1900.04.09

1902.09.26

1927.05.18

2173.10.14

{ } is a function with argument x

0 60?x index of x among 0 60 or 2 if not found

ˋ1900.01.00ˋ1900.02.29 a list of two symbols

,"d"$x-36526-x<60 append to it a date x days after 1900.01.00, adjusted for excel's leap error

(ˋ1900.01.00ˋ1900.02.29,"d"$x-36526-x<60)@0 60?x juxtaposition means indexing - the "@" in the middle is implicit

$ convert to string