Let $\tau = \min \{ n \geq 1:X_1 + \cdots + X_n > K \}$. Then $\tau$ is an integer-valued random variable, bounded from above by $K+1$ (since $X_i \geq 1$). Note that $\tau = n$ if and only if $\sum\nolimits_{i = 1}^{n - 1} {X_i } \le K$ and $\sum\nolimits_{i = 1}^{n} {X_i } > K$. Thus, the event $\lbrace \tau = n \rbrace$ depends only on the values $X_1,\ldots,X_n$. So, by definition, $\tau$ is a stopping time with respect to the sequence $X_1,X_2,\ldots$. Now, $X_1,X_2,\ldots$ are i.i.d. with finite expectation $\mu$, and $\tau$ is a stopping time for them. Moreover, ${\rm E}(\tau) < \infty$ since $\tau \leq K+1$. Hence, by Wald's identity,
${\rm E}\bigg(\sum\limits_{i = 1}^\tau {X_i } \bigg) = {\rm E}(\tau )\mu \leq (K+1)\mu.
$
So if we put $Y_\tau = \sum\nolimits_{i = 1}^\tau {X_i }$, we get
${\rm E}(Y_\tau - K) = {\rm E}(Y_\tau) - K \leq (K+1)\mu - K.$

EDIT: Since $\tau \geq 1$, we have
$\mu - K \leq {\rm E}(Y_\tau - K) \leq (K+1)\mu - K. $

As we have seen above, the problem reduces to calculating ${\rm E}(\tau)$. Put $S_n = \sum\nolimits_{i = 1}^n {X_i }$ ($S_0 = 0$).
Note that
$ {\rm P}(\tau = n) = {\rm P}(S_{n - 1} \le K,S_n > K) = {\rm P}(S_{n - 1} \le K) - {\rm P}(S_n \le K). $
Hence, $ {\rm E}(\tau) = \sum\limits_{n = 1}^{K + 1} {n{\rm P}(\tau = n)} = \sum\limits_{n = 1}^{K + 1} n[{\rm P}(S_{n - 1} \le K) - {\rm P}(S_n \le K)] = \sum\limits_{n = 0}^K {{\rm P}(S_n \le K)}. $
So, we can write
$ {\rm E}(\tau) = 1 + \sum\limits_{n = 1}^\infty {{\rm P}(S_n \le K)} = 1 + \sum\limits_{n = 1}^\infty {F^{(n)}(K)}, $
where $F^{(n)}$ is the distribution function of $S_n$. For $t>0$ real, define $m(t) = \sum\nolimits_{n = 1}^\infty {F^{(n)}(t)}$.
From the theory of renewal processes, we know that $m(t) = {\rm E}(N_t)$, where $\lbrace N_t:t \geq 0 \rbrace$ is a renewal process with inter-arrival times distributed according to the distribution of $X$. $m(t)$ is called the renewal function. It may be worth noting that by the Elementary Renewal Theorem,
$\mathop {\lim }\limits_{t \to \infty } \frac{{m(t)}}{t} = \frac{1}{\mu }. $
Returning to our original setting, we have
$ {\rm E}(Y_\tau ) = {\rm E}(\tau )\mu = (1 + m(K))\mu. $
So, the problem reduces to calculating $m(K)$.

EDIT: Elaborating on the relation to the framework of renewal theory.

For completeness and for general purposes, let us consider the problem in the (more general) setting of renewal theory. For this purpose, we replace $Y$ by $S$, in accordance with the common notation used in renewal theory. Henceforth we suppose that $X_i$ are i.i.d. non-negative rv's with mean $\mu > 0$, and set $S_n = \sum\nolimits_{i = 1}^n {X_i }$. For $t \geq 0$ real, we set $\tau_t = \inf \{ n : S_n > t \}$ (thus further generalizing the case considered in the question). We now introduce the stochastic process $N = \lbrace N_t : t \geq 0 \rbrace$, defined by $N_t = \sup \{ n:S_n \le t\}$. The counting process $N$ is called a renewal process. The key observation is that $\tau_t$ and $N_t$ are related by $\tau_t = N_t + 1$. (Note that thus $N_t + 1$ is a stopping time for the $X_i$.) Thus, $S_{\tau _t } = S_{N_t + 1}$. This corresponds to $Y_\tau$ of the original question, upon letting $t=K$. However, in accordance with the common notation used in renewal theory, we shall use $Y$ for the following random variable: we define $Y_t = S_{N_t + 1} - t$. The random variable $Y_t$ is called the excess at $t$ of the renewal process $N$. Thus, $Y_t = S_{\tau _t } - t$, and so (by letting $t=K$) this corresponds to the random variable denoted $Y_\tau - K$ in the original question. Hence, as it turns out, the OP actually considered the expectation of the excess at $K$ of a renewal process with inter-arrival times distributed according to the distribution of the $X_i$.

Finally, here is some useful link concerning renewal theory, which is very relevant to this answer.