Does there exist positive rational $s$ for which $\zeta(s)$ is a positive integer? 
Does there exist positive rational $s$ for which the Riemann Zeta function $\zeta(s) \in N$ or equivalently, does there exist finite positive integers $\ell,m$ and $n$ such that $$\zeta\left(1+\dfrac{\ell}{m}\right) = n$$

Update: 16-Apr-2021: New and faster code using hash map. Using this code and the method described in my answer below, I have been able to show that if there is a solution then $l > 1.7\times 10^5$.
def get_c_value(m):
    i   = 1
    c_m = 0
    while (i <= r):
        c_m = N(m + c_m - zeta(1 + 1/(m - 1 + c_m)), digits = max_digits)
        i = i + 1
    return c_m

def n_max_skip(a,b):
    n_skip = 1
    while b <= b_max:
        if gcd(b,a) == 1:
            c_b = N(b/a, digits = max_digits)
            n_b = 1 + floor(0.0728158454836767248605863758749013191377363383/(eg - c_b))
            if n_b > n_skip + 1:
                return n_skip
            n_skip = n_b
        b = b + 1  

r          = 10     # Number of iterations for computing estimates
max_digits = 55     # All estimate will have an accuracy of 50 digits after the decimal point
eg         = N((1 - euler_gamma), digits = max_digits)
c_dict     = {}
a          = 1
n_max      = 0
count_max  = 0

while a <= 10^9:
    b_min  = 1 + floor(0.372406215900714*a)
    b_max  = floor(eg*a)
    b      = b_min
    n      = 2
    n_skip = n_max_skip(a,b_min)
    
    while n <= n_skip:
        if n in c_dict:
            c = c_dict.get(n)
        else:
            c = get_c_value(n)
            c_dict[n] = c
            
        b = floor(c*a)
        
        if not (N(b/a, digits = max_digits) < c and c < N((b + 1)/a, digits = max_digits)):
            print('Exception',a,b,n)
            break
        n = n + 1
        
    b = b + 1
    
    while b <= b_max:
        if(gcd(b,a) == 1):
            test    = N(b/a, digits = max_digits)
            n       = max(2,floor(0.0728158454836767248605863758749013191377363383/(eg - test)))
            n_start = n
            c_prev  = get_c_value(n)
            not_found =  1
            
            while not_found == 1:
                c_n1 = get_c_value(n + 1)          
                if c_prev < test and test < c_n1:
                    not_found = 0
                    if  n > n_max:
                        n_max = n
                        count_max = count_max + 1
                        print("Maximum n is at:", a, b, n_max, count_max)
                else:
                    c_prev = c_n1
                n = n + 1
        b = b + 1
    #print(a)
    a = a + 1

 A: Whilst not a solution, I thought it might be interesting to see What happens if we use the functional equation:
$$\zeta\left(1+\frac{m}{n}\right)=2^{\left(1+\frac{m}{n}\right)}\pi^{\frac{m}{n}}\sin\left(\frac{\pi}{2}\left(1+\frac{m}{n}\right)\right)\Gamma\left(-\frac{m}{n}\right)\zeta\left(-\frac{m}{n}\right).$$
$\zeta(s)$ is known to be rational at the negative integers,
$$\zeta(-a)=(-1)^a\frac{B_{a+1}}{a+1}.$$
You actually get $\zeta(-a)=0$ for $a$ even due to the trivial zeros.
But it appears that when you combine this with the functional equation, then in the limit, you get something non-zero and irrational, so that doesn't even give you an integer.
On the other hand if $m/n=2k-1$ is odd then we have
$$\zeta\left(2k\right)=\frac{(-1)^{k+1}B_{2k}(2\pi)^{2k}}{2(2k)!},$$
by Euler's formula, which is "almost" rational except for the $\pi$ factor. Bugger. So no chance of being integral.
If you modify your problem slightly, then using the above you can produce positive integers, but that's no fun.
A: I misread $l$ as $1$, but in any case, as a partial result, here's a resolution for the case $l=1$.

Fix $s\in\mathbb{R}$, with $s > 1$.

On the interval $(0,\infty)$, let 
$f(x)={\small{{\displaystyle{\frac{1}{x^{\large{s}}}}}}}$.

It's easily verified that 
$
{\displaystyle{
\int_{1}^\infty \!f(x)\,dx
=
{\small{\frac{1}{s-1}}}
}}
$.

Consider the infinite series 
$
{\displaystyle{
\sum_{k=1}^\infty \frac{1}{k^s}
}}
$.

Since $f$ is positive, continuous, and strictly decreasing, we get
\begin{align*}
\int_{1}^\infty \!f(x)\,dx < \;&\sum_{k=1}^\infty \frac{1}{k^s} < 1+\int_{1}^\infty \!f(x)\,dx\\[4pt]
\implies\;{\small{\frac{1}{s-1}}} < \;&\sum_{k=1}^\infty \frac{1}{k^s} < 1+{\small{\frac{1}{s-1}}}\\[4pt]
\end{align*}
If $m$ is a positive integer, then letting $s=1+{\large{\frac{1}{m}}}$, we have ${\large{\frac{1}{s-1}}}=m$, hence
\begin{align*}
{\small{\frac{1}{s-1}}} < \;&\sum_{k=1}^\infty \frac{1}{k^s} < \;1+{\small{\frac{1}{s-1}}}\\[4pt]
\implies\;m < \;\,&\zeta\bigl(1+{\small{\frac{1}{m}}}\bigr) < \;m + 1\\[4pt] 
\end{align*}
so  $\zeta\bigl(1+{\large{\frac{1}{m}}}\bigr)$ is not an integer.
A: 
Can you resolve the problem for any other value of l, other than l=1?
For example, can you resolve the case l=2?

Yes and in fact I can show that there is no solution for $l \le 3*10^4$. Here is the outline of my approach which I am posting as an answer since it is too long to be a comment.
Step 1: The first step was to derive the following result

For every real $x \ge 1$ there exists a positive real $c_x$ such
that ${\displaystyle{ \zeta\Big(1+\frac{1}{x-1+c_x}\Big) = x. }}$
The first few terms of the asymptotic expansion of $c_x$ in terms of $n$ and the Stieltjes constants $\gamma_i$ are
$$ c_x = 1-\gamma_0 + \frac{\gamma_1}{x-1} 
 + \frac{\gamma_2 + \gamma_1 - \gamma_0 \gamma_1}{(x-1)^2}
+ \frac{\gamma_2 +2\gamma_2 - 2\gamma_2 \gamma_0 + \gamma_1 - 2\gamma_1 \gamma_0 + \gamma_1 \gamma_0^2 - \gamma_1^2}{(x-1)^3}
+ O\Big(\frac{1}{x^4}\Big) $$

Step 2: I computed the first few values of $c_n$ but I did not use the above result. Instead I used the following recurrence formula.

Let $\alpha_0$ be any positive real and  ${\displaystyle{ \alpha_{r+1} = n +
\alpha_r - \zeta\Big(1+\frac{1}{n -1 + \alpha_r}\Big); }}$ then ${\displaystyle{ \lim_{r \to \infty}\alpha_r = c_n}}$.

Using this we obtained
$$
c_2 \approx 0.3724062
$$
$$
c_3 \approx 0.3932265
$$
$$
\ldots
$$
$$
c_{12} \approx 0.4164435
$$
Step 3: Show that $l \ge 5$
Let ${\displaystyle{ \zeta\Big(1+\frac{l}{m}\Big) \in N}}$ and let $m = lk+d$ where $\gcd(l,d) = 1$ and $1 \le d < l$.
Clearly, $c_2 \le c_n < 1-\gamma_0$ or $0.3724062 \le c_n < 0.422785$.
Hence we must have ${\displaystyle{ 0.3724062 \le \frac{d}{l} < 0.422785}}$. The fraction with the smallest value of $l$ satisfying this condition is ${\displaystyle{\frac{2}{5} }}$ hence $l \ge 5$.
Extending this approach using numerical computations, I am able to show that there is no solution for $l < 3*10^4$.
Problems with this approach:
With this approach and with powerful computing, we can prove results like if ${\displaystyle{  \zeta\Big(1+\frac{l}{m}\Big) \in N }}$ then $l$ must be greater than some large positive integer but I don't see how this approach will solve the general problem.
