If the mean of four of six given numbers is known, what is the mean of the other two?

Suppose $a,b,c,d$ are the four numbers whose mean is 2008. That means that $$\frac{a+b+c+d}{4} = 2008.$$

If $e$ and $f$ are the other two numbers, you want to find $$\frac{e+f}{2}.$$

Well, if you know $e+f$, you can figure out their mean (just divide by $2$). And you happen to know how much the sum of all six numbers is (you can just compute it). Can you use the information you have to figure out what $e+f$ must be?

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Added. Given the comment by the OP below, it seems the hint above is insufficient.

You don't need to know the four numbers. Nobody is asking you what the four numbers are, or what the other two numbers are. All you need to know is what $e+f$ is. Because once you know what the sum of "the other two numbers" is, then you can just divide that sum by $2$ and get the mean.

Now, you know what $a+b+c+d+e+f$ (the sum of all six numbers) is: it's $$1867 + 1993 + 2019 + 2025 + 2109 + 2121 =12134.$$ (I'm not saying that $a$ is 1867; I'm just saying that adding all six numbers is 12134).

You also know what the sum of the four numbers whose mean is 2008 is: because you know that $(a+b+c+d)/4 = 2008$, so that means that the sum of the four numbers is $4\times 2008 = 8032$.

So: the sum of all six numbers is $12134$. The sum of the four numbers with mean 2008 is $8032$.

How much is the sum of "the other two numbers"?

And if the sum of the other two numbers it that much, how much is their mean?

Hint: If $X$ is the mean of $N$ numbers then their sum is...?