TITLE: Does there exist positive rational $s$ for which $\zeta(s)$ is a positive integer?
QUESTION [30 upvotes]: 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

REPLY [5 votes]: 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.