Let's denote Dan's share of the day's work as \( D \) and the total work that the team has to complete for the day as \( T \).
From the problem, we know that Dan has completed \( \frac{1}{3} \) of his share of the work, so we can express this as:
\[ \text{Work completed by Dan} = \frac{1}{3} D \]
It is also given that Dan's share of the work is at least \( \frac{1}{4} \) of his team's total work, which can be expressed mathematically as:
\[ D \geq \frac{1}{4} T \]
Next, we need to find out what portion of the total work \( T \) Dan has completed.
We can substitute \( D \) with \( \frac{1}{4}T \) (since that's the minimum share) to find out the smallest contribution to the total work:
\[ \text{If } D = \frac{1}{4} T, \text{ then } \frac{1}{3} D = \frac{1}{3} \cdot \frac{1}{4} T = \frac{1}{12} T \]
So under the minimum condition where Dan's share is exactly \( \frac{1}{4}T \), he has completed \( \frac{1}{12}T \) of the total work.
Since \( D \) could be greater than \( \frac{1}{4}T \), it could yield a greater amount of work completed. To express this in general terms if we let \( D \) be any value greater than or equal to \( \frac{1}{4} T \):
If \( D = kT \) where \( k \geq \frac{1}{4} \), then Dan's contribution would be:
\[ \text{Work completed by Dan} = \frac{1}{3} D = \frac{1}{3} (kT) = \frac{k}{3} T \]
Thus, the portion of the total amount of work Dan has completed is given by:
\[ \frac{1}{3} D \text{ where } D \geq \frac{1}{4} T \]
In the minimum case, the portion of work Dan has completed is \( \frac{1}{12} \) of \( T \), and it can increase based on the actual value of \( D \), but it will always be \( \geq \frac{1}{12}T \) (when \( D \geq \frac{1}{4}T \)).
In conclusion, the portion of the total work Dan has completed is:
\[ \text{At least } \frac{1}{12} \text{ of the total work } T. \]