The required integral part is $0$. The difference of the two sums is positive, since positivity clearly holds termwise.

To show it less than $1$, let $u_k = n^k/k!$, $k = 0, 1, 2, \dots$, and express the two sums in terms of the $u_k$. The first sum is $(u_{n+1} + \cdots + u_{2n-1})/u_n$, and the second is $(u_0 + \cdots + u_{n-2})/u_{n-1}$. Notice that $u_{n-1} = u_n$ to write the difference of the two in the form
$$1 - \frac{u_{2n}}{u_n} + \frac{1}{u_n} \sum_{k=1}^{n} (u_{n+k} - u_{n-k}).$$
It is therefore sufficient to prove that $\sum_{k=1}^n u_{n+k} < \sum_{k=1}^n u_{n-k}$. To this end, we will show that
$$\sum_{k=1}^{n} k u_{n+k} < \sum_{k=1}^{n} k u_{n-k} \quad \text{and} \quad \sum_{k=1}^{n} (m-k) u_{n+k} < \sum_{k=1}^{n} (m-k) u_{n-k} \quad (*)$$
for some $m$ in the range $1, 2, \dots, n$. Addition of the two inequalities $(*)$ yields the desired inequality.

The first inequality $(*)$ is equivalent to $\sum_{k=0}^{2n} (k-n) u_k < 0$. To establish the latter, simply notice that $\sum_{k=0}^{2n} k u_k = n \sum_{k=0}^{2n-1} u_k < n \sum_{k=0}^{2n} u_k$.

To establish the second inequality $(*)$, discard the trivial case $n = 2$ and let $n \ge 3$. Write
$$\frac{u_{n+k}}{u_{n-k}} \cdot \frac{u_{n-k-1}}{u_{n+k+1}} = 1 + \frac{1}{n} - \frac{k(k+1)}{n^2}, \quad k = 0, 1, \dots, n,$$
to infer that, as a function of $k$, the ratio $u_{n+k}/u_{n-k}$ is strictly decreasing for $k(k+1) < n$ and strictly increasing for $k(k+1) > n$; the situation at the possible case $k(k+1) = n$ causes no trouble. Since $u_{n+1}/u_{n-1} = n/(n+1) < 1$, and $u_{2n}/u_0 = n^{2n}/(2n)! > 1$ for $n \ge 3$, there exists $m$ in the range $1, 2, \dots, n$ such that $u_{n+k} < u_{n-k}$ for $1 \le k < m$, and $u_{n+k} > u_{n-k}$ for $m < k \le n$. Consequently, the second inequality $(*)$ holds termwise for $k \ne m$. This ends the proof.