Let $f : \mathbb{Q} \to \mathbb{Q}$ be a function such that for any $x, y \in \mathbb{Q}$, the number $f(x+y) - f(x) - f(y)$ is an integer. We need to determine whether there exists a constant $c$ such that $f(x) - cx$ is an integer for every rational number $x$.

To address this, we construct a counter-example. We begin by recursively constructing a sequence $k_1, k_2, \ldots$ with $0 \leq k_q < q$ for each $q$, as follows:
1. $k_1 = 0$.
2. If $q > 1$ and $k_1, \ldots, k_{q-1}$ are already defined, choose $k_q$ such that $0 \leq k_q < q$ and $k_q \equiv k_d \pmod{d}$ whenever a positive integer $d$ divides $q$.
3. If $q$ is a prime, set $k_q = 1$ if $q = 2$ or $q \equiv 1 \pmod{73}$, and $k_q = 2$ otherwise.

This construction is well-defined by induction. For the base case, $k_1 = 0$ is trivial. For the induction step:
- If $q$ is not a prime, the Chinese Remainder Theorem ensures the existence of such a $k_q$ if $\gcd(d_1, d_2) \mid k_{d_1} - k_{d_2}$ for any two distinct proper divisors $d_1, d_2$ of $q$. From the induction hypothesis, $k_{d_1} \equiv k_d \pmod{d}$ and $k_{d_2} \equiv k_d \pmod{d}$, implying $k_{d_1} \equiv k_{d_2} \pmod{d}$.
- If $q$ is a prime, the congruences $k_q \equiv k_1 \pmod{1}$ and $k_q \equiv k_q \pmod{q}$ are trivially satisfied.

Next, we define $f$ for all rationals as $f\left(\frac{p}{q}\right) = \left\{\frac{pk_q}{q}\right\}$, where $\{x\}$ denotes the fractional part of $x$. This definition is well-defined because $\left\{\frac{pk_q}{q}\right\} = \left\{\frac{pnk_{qn}}{qn}\right\} = \left\{\frac{pk_{qn}}{q}\right\}$, and $k_{qn} \equiv k_q \pmod{q}$ implies $pk_{qn} \equiv pk_q \pmod{q}$.

Given two rationals $x$ and $y$, choose a common denominator $r$ and write $x = \frac{X}{r}$ and $y = \frac{Y}{r}$. Then $f(x+y) = \left\{\frac{(X+Y)k_r}{r}\right\}$ and $f(x) + f(y) = \left\{\frac{Xk_r}{r}\right\} + \left\{\frac{Yk_r}{r}\right\}$, and the difference between these numbers is an integer.

Assume there is a constant $c$ such that $f(x) - cx$ is an integer for every rational $x$. Since $f(1) = 0$, $c \in \mathbb{Z}$. For every prime $p$, $\frac{k_p - c}{p}$ must be an integer. Since there are infinitely many primes $\equiv 1 \pmod{73}$, $c = 1$, and since there are infinitely many primes not congruent to $1 \pmod{73}$, $c = 2$, leading to a contradiction.

Therefore, there does not exist a constant $c$ such that $f(x) - cx$ is an integer for every rational number $x$. The answer is: \boxed{\text{No}}.