First of all, $\sqrt{2}+\sqrt{3}$ most certainly is a number. It is a real number, approximately equal to $3.14626$ Perhaps what you're asking is why the sum of two simple radicals isn't also a simple radical, when the sum of two integers is an integer, and the sum of two fractions is a fraction.
Roots of integers are examples of algebraic numbers - numbers that are the roots of polynomial equations with integer coefficients. The sum of two algebraic numbers is an algebraic number, but it doesn't have to be as simple as the addends are.
Something like this is already true with fractions: $\frac12$ and $\frac13$ are both unit fractions, but their sum is a more complicated fraction, with larger numbers in both the numerator and denominator than the fractions we started with.
Similarly, the numbers $\sqrt2$ and $\sqrt3$ are roots of the polynomials $x^2-2$ and $x^2-3$, respectively. Their sum, which is most simply expressed as $\sqrt2+\sqrt3$, is a root of the polynomial $x^4-10x^2+1$.
This particular algebraic number can also be written as $\sqrt{5+2\sqrt{6}}$, where you can see both $2+3$ and $2\times 3$ playing a role. (To see this, write $x=\sqrt2+\sqrt3$, square both sides, and combine integers.)
Just like you can't expect reciprocals of integers to be closed under addition, you can't expect roots of integers to be closed under addition. That's because we're involving more complicated operations than addition, namely, division and taking of roots. On the other hand, you can expect the sum of two rational numbers (fractions) to be rational, and you can expect the sum of two algebraic numbers (roots of integer polynomials) to be an algebraic number.
Does this help at all?
As requested in the comments, I'll put this in the voice I would use to address a high school or middle school student asking this question:
"First of all, $\sqrt2 + \sqrt3$ is definitely a number. Here, let's calculate its value... [calculator].... as you can see, it's not a very pretty one, but we can see a few decimal places: $3.14626\ldots$. Huh, it's kind of close to $\pi$, but a little bigger.
"Anyway, let's see if we can express this number in a nicer form:
$$\begin{align}
x &= \sqrt2 + \sqrt3\\
x^2 &= (\sqrt2 + \sqrt3)^2\\
x^2 &= 2 + 2\sqrt2\sqrt3 + 3\\
x^2 &= 5 + 2\sqrt6\\
x &= \sqrt{5+2\sqrt6}
\end{align}$$
(I'd talk through the steps of that algebra, making sure it's clear after each line.)
"Ok, so it's a square root, but it's a square root of something more complicated that what we started with. I guess that's fair. After all, when you add two fractions like $\frac12$ and $\frac13$, which are pretty simple, you end up with $\frac56$, which is more complicated - it's not just a $1$ on top, and both numerator and denominator are bigger than what we started with.
"Actually, it's pretty interesting, that the numbers $5$ and $6$, which are $2+3$ and $2\times 3$, both show up in the fraction $\frac56$ and in the radical $\sqrt{5+2\sqrt{6}}$
"The reason adding fractions is more complicated than adding integers, and adding radicals more complicated still, is that fractions are made of division, and radicals are made of roots, both of which are more complicated than addition and subtraction in the first place."
Second edit:
One more run at this, just to see how succinctly I can get the main point.
"To see what something is the square root of, square it:
$$(\sqrt2 + \sqrt3)^2 = 2 + 2\sqrt6 + 3 = 5 + 2\sqrt6$$
"As you can see, we don't get a whole number, because in FOIL*, we have middle terms giving us the $2\sqrt6$ part.
"It's different from a fraction, because if you look at the sum $\frac12 + \frac13$, there is a common demonimator $6$ you can multiply by that makes it the sum of two whole numbers: another whole number. No cross terms arise, because there's no FOIL going on."
(* FOIL = distributive rule applied to binomials; mnemonic for "First terms, Outside terms, Inside terms, Last terms")
Also instructive are the cases where it does work. For example, $\sqrt2 + \sqrt8 = \sqrt{18}$. You can "FOIL" it out and see why.