tl;dr: You might spend less time wondering about peoples' sick days. :-)
It depends a lot on what you consider suspicious, and also (possibly) how many employees you're keeping tabs on.
In the most restrictive case, you only consider Monday suspicious, and are only keeping tabs on this one employee. In that case, the probability (given the null hypothesis that the employee is equally likely to be sick on any given day of the week) that at least four of seven sick days fall on a Monday is
\begin{align}
P & = \binom{7}{4} \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^3 \\
& + \binom{7}{5} \left(\frac{1}{5}\right)^5 \left(\frac{4}{5}\right)^2 \\
& + \binom{7}{6} \left(\frac{1}{5}\right)^6 \left(\frac{4}{5}\right)^1 \\
& + \binom{7}{7} \left(\frac{1}{5}\right)^7 \left(\frac{4}{5}\right)^0 \\
& = \frac{2240+336+28+1}{78125} \\
& = \frac{2605}{78125} = \frac{521}{15625} \doteq 0.033344
\end{align}
or about $1$ chance in $30$. Just to be clear, that is not the probability that the employee is malingering. It is the probability, given that the employee is not malingering, that they would exhibit this sort of sick-day record. This level of result is generally not considered statistically significant.
And note that this is given that you're only looking at one employee. If you work in an office with ten employees, say, it's far more likely that you'll notice one employee with this kind of record, purely by dumb luck, even if none of them is malingering. The probability there is given by
$$
P = 1 - \left(1-\frac{521}{15625}\right)^{10} \doteq 0.28761
$$
or over a $1$ in $4$ chance that at least one of the employees will be out at least four Mondays out of seven sick days. It's really not that unusual.
And furthermore, this assumes that only Mondays are suspicious. Why shouldn't Fridays be suspicious, too? That raises the odds of a completely innocent trend correspondingly.
It is of course unrealistic that the chance of a person born on a Friday is $\frac{1}{3}$, but so be it. We are also told that for all other days of the week the chance is the same, so that would be
$$\frac{1-\frac{1}{3}}{6}=\frac{1}{9}$$ (e.g. the chance of being born on a Wednesday is $\frac{1}{9}$)
Now, there are various ways to have 2 people being born on the same day, and the other two on 2 other days:
$A$. Two born on a Friday, and the other two on two different days other than Friday
$B$. Two born on the same day but other than a Friday, one born on Friday, and the last on a different day yet.
$C$. Two born on the same day but other than a Friday, and the other two on two different days other than Friday
$$P(A) = {4 \choose 2} \cdot \big( \frac{1}{3} \big)^2\cdot \frac{6}{9} \cdot \frac{5}{9} = \frac{180}{729}$$
(the $4 \choose 2$ is the number of ways to pick the two people born on a Friday)
$$P(B) = {4 \choose 2} \cdot 6 \cdot \big( \frac{1}{9} \big)^2\cdot 2 \cdot \frac{1}{3} \cdot \frac{5}{9} = \frac{120}{729}$$
(the extra $2$ term is to distinguish the two people: one born on Friday, the other not. The $6$ is to pick one of the 6 days that the first two are born on)
$$P(C) = {4 \choose 2} \cdot 6 \cdot \big( \frac{1}{9} \big)^2 \cdot \frac{5}{9} \cdot \frac{4}{9} = \frac{80}{729}$$
Hence, the chance that 2 people being born on the same day, and the other two on 2 other days is:
$$P(A)+P(B)+P(C)=\frac{380}{729} \approx 0.52$$
Best Answer
Since you need at least a number in $\{2,4,6\}$ and a number in $\{5\}$ then you can count the probability to never obtain a number in the two sets or to obtain just one of them out of $n$ trials using the binomial distribution
$$P(\text{none}) =( \frac{3}{7})^n$$ $$P(\text{not both [case 1]}) =\sum_{k=1}^{n}{n \choose k} (\frac{3}{7})^n$$ $$P(\text{not both [case 2]}) =\sum_{k=1}^{n}{n \choose k} (\frac{1}{7})^k(\frac{3}{7})^{n-k}$$
So the probability is
$$p =1-( \frac{3}{7})^n - \sum_{k=1}^{n}{n \choose k} (\frac{3}{7})^n-\sum_{k=1}^{n}{n \choose k} (\frac{1}{7})^k(\frac{3}{7})^{n-k} \iff $$
$$p-(\frac37)^n-(\frac37)^n = 1-( \frac{3}{7})^n - (\frac{3}{7}+\frac{3}{7})^n- (\frac{1}{7}+\frac{3}{7})^{n} \iff $$ $$p = 1 +(\frac37)^n - (\frac{6}{7})^n- (\frac{4}{7})^{n} $$
You want $p>0.5 \iff (\frac{6}{7})^n+ (\frac{4}{7})^{n} -(\frac{3}{7})^n < 0.5 \iff 6^n+4^n-3^n < \frac12 \cdot 7^n$
The two functions (on the left and right side of the inequality) are increasing at rates so that when $n$ is positive they have just one point of intersection and after that the rhs function is always greater than the other, therefore it's not difficult through experimentation to see that $n = 6$ is the required amount of people.