Encouraged by the comment of Jean Marie, i will slightly enlarge my comment and post it so as an answer. Let us fix a natural number $N$, and after excluding the trivial case $N=1$ we may and do assume that $N\ge 2$. Let $P$ be the list of primes less or equal $N$. Then the given number
$$\xi=\sum_{1\le k\le N}\sqrt k$$
lives in the field $K=\Bbb Q(\sqrt p\ :\ p\in P)$. This field is realized naturally as a tower of fields, involving extensions of degree two, at each step we adjoin the square root of a prime in $P$. For instance, using the natural order of $P$ to adjoin,
$$
\Bbb Q\to
\Bbb Q(\sqrt 2)\to
\Bbb Q(\sqrt 2,\sqrt 3)\to\dots\to
\Bbb Q(\sqrt 2,\sqrt 3, \dots,\sqrt{\max P})=K\ .
$$
Then the Galois group of the Galois extension $K$ of $\Bbb Q$ splits as a product of $|P|$ copies of groups with two elements, each generated by a Galois morphism of the shape $\sqrt p\to-\sqrt p$. So the minimal polynomial of $\xi$ is
$$
f_N(X)=\prod_{e:P\to\{-1, 1\}}\left(\ X - \left(\sum_{1\le k\le N} \bar e(k)\sqrt k\right)\ \right)\ .
$$
Here, $\bar e$ is the unique multiplicative extension of $e$, which is defined only on $P$, to all natural numbers $k$ from $1$ to $N$. To get $\bar e(k)$ we factor $k$ over the integers, in the factorization there are only primes from $P$, taken to some powers, and we define $\bar e(k)$ to be the sign obtained from the factorization by replacing each prime $p$ by the sign $e(p)$. For instance, if the following sample primes are in $P$, $\bar e(4)
=\bar e(2^2)=e(2)^2=1$, and it does not depend on $e$, same argument gives $e(k^2)=1$, $\bar e(6)=\bar e(2\cdot 3):=e(2)\cdot e(3)$, $\bar e(12)=\bar e(2^2\cdot 3):=e(2)^2\cdot e(3)=e(3)$, and so on.
The formula is not really useful, but it shows the the degree of $f_N$ is $2^{|P|}$. For instance, if $N=6$, then there $P=\{2,3,5\}$, and we expect the degree $2^{|P|}=2^3=8$ for this minimal polynomial. Indeed, asking for this polynomial in sage...
sage: xi = (sum([sqrt(k) for k in [1..6]])).minpoly()
sage: xi
x^8 - 24*x^7 + 188*x^6 - 456*x^5 - 722*x^4 + 3768*x^3 - 468*x^2 - 6552*x + 441
we get a polynomial of degree eight. For $N=7$ there is a jump, the degree is $2^4=16$, and this is so for $N=7,8,9,10$, the next jump in degree being for $N=11$, with degree $2^5=32$.
Explicitly, the above polynomial $f_6$ was obtained as a product of the following linear factors:
$$
\begin{aligned}
&X - (1 +\sqrt 2+\sqrt 3 + 2 +\sqrt 5 +\sqrt 6)\ ,& e=(+1,+1,+1)\ ,\\
&X - (1 +\sqrt 2+\sqrt 3 + 2 -\sqrt 5 +\sqrt 6)\ ,& e=(+1,+1,-1)\ ,\\[2mm]
&X - (1 +\sqrt 2-\sqrt 3 + 2 +\sqrt 5 -\sqrt 6)\ ,& e=(+1,-1,+1)\ ,\\
&X - (1 +\sqrt 2-\sqrt 3 + 2 -\sqrt 5 -\sqrt 6)\ ,& e=(+1,-1,-1)\ ,\\[4mm]
&X - (1 -\sqrt 2+\sqrt 3 + 2 +\sqrt 5 -\sqrt 6)\ ,& e=(-1,+1,+1)\ ,\\
&X - (1 -\sqrt 2+\sqrt 3 + 2 -\sqrt 5 -\sqrt 6)\ ,& e=(-1,+1,-1)\ ,\\[2mm]
&X - (1 -\sqrt 2-\sqrt 3 + 2 +\sqrt 5 +\sqrt 6)\ ,& e=(-1,-1,+1)\ ,\\
&X - (1 -\sqrt 2-\sqrt 3 + 2 -\sqrt 5 +\sqrt 6)\ ,& e=(-1,-1,-1)\ .
\end{aligned}
$$
Well, sage may have done something else, so let us check explicitly the result obtained by multiplying these linear factors:
sage: var('X');
sage: prod([ X - (1 + e2*sqrt(2) + e3*sqrt(3) + 2 + e5*sqrt(5) + e2*e3*sqrt(6))
....: for (e2, e3, e5) in cartesian_product([[-1, 1], [-1, 1], [-1, 1]]) ]) \
....: .expand().canonicalize_radical()
X^8 - 24*X^7 + 188*X^6 - 456*X^5 - 722*X^4 + 3768*X^3 - 468*X^2 - 6552*X + 441
sage:
Et voilà!
