Which Day of Christmas is it?

Python, 27 bytes

lambda n:int((n*6)**.33359)

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A direct formula with some curve-fitting, same as the original one found by Level River St.

The shifted equation i**3-i==n*6 is close to i**3==n*6 for large i. It solves to i=(n*6)**(1/3). Taking the floor the rounds down as needed, compensating for the off-by-one.

But, there are 6 inputs on boundaries where the error takes it below an integer it should be above. All of these can be fixed by slightly increasing the exponent without introducing further errors.

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Python, 38 bytes

f=lambda n,i=1:i**3-i<n*6and-~f(n,i+1)

The formula n=i*(i+1)*(i+2)/6 for tetrahedral numbers can be more nicely written in i+1 as n*6=(i+1)**3-(i+1). So, we find the lowest i for which i**3-i<n*6. Each time we increment i starting from 1, the recursive calls adds 1 to the output. Starting from i=1 rather than i=0 compensates for the shift.

J, 12 bytes

2>.@-~3!inv]

There might be a golfier way to do this, but this is a lovely opportunity to use J's built-in function inversion.

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How it works

2>.@-~3!inv]  Monadic verb. Argument: n

           ]  Right argument; yield n.
      3       Yield 3.
       !inv   Apply the inverse of the ! verb to n and 3. This yields a real number.
              x!y computes Π(y)/(Π(y-x)Π(x)), where Π is the extnsion of the 
              factorial function to the real numbers. When x and y are non-negative
              integers, this equals yCx, the x-combinations of a set of order y.
 >.@-~        Combine the ceil verb (>.) atop (@) the subtraction verb (-) with
              swapped arguments (~).
2             Call it the combined verbs on the previous result and 2.

Python, 22 bytes

lambda n:n**.3335//.55

Heavily inspired by @xnor's Python answer.

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