First, let $a=1$. It holds that $\varphi(1)=1$. For all positive integers $b$, we have $\operatorname{ggd}(a, b)=1$, so the equation becomes $1+b=1+\varphi(b)+1$, thus $\varphi(b)=b-1$. Therefore, there is exactly one number in $\{1,2, \ldots, b\}$ that is not coprime with $b$; this must be $b$ itself (since $\operatorname{ggd}(b, b)>1$ unless $b=1$, but in that case $\varphi(b)=b$). This implies that $b$ is a prime number. The equation holds for all prime numbers. We conclude that $(1, p)$ is a solution for all prime numbers $p$. Similarly, $b=1$ gives the solutions $(p, 1)$.

Now, we assume that $a, b \geq 2$. Since $\operatorname{ggd}(b, b)>1$, it follows that $\varphi(b) \leq b-1$. Therefore,

$$\operatorname{ggd}(a, b)=a+b-\varphi(a)-\varphi(b) \geq a-\varphi(a)+1$$

Let $p$ be the smallest prime divisor of $a$ (which exists, since $a \geq 2$). Since for all multiples of $p$, $t p \leq a$, we have $\operatorname{ggd}(t p, a)>1$, it follows that $a-\varphi(a) \geq \frac{1}{p} \cdot a$. Therefore,

$$\operatorname{ggd}(a, b) \geq a-\varphi(a)+1 \geq \frac{a}{p}+1$$

The largest two divisors of $a$ are $a$ and $\frac{a}{p}$. Since $\operatorname{ggd}(a, b)$ is a divisor of $a$ that is at least $\frac{a}{p}+1$, it must be equal to $a$. Thus, $\operatorname{ggd}(a, b)=a$. By a completely analogous argument, we can prove that $\operatorname{ggd}(a, b)=b$. Therefore, $a=b$.

Now the equation becomes $2 a=2 \varphi(a)+a$, or $a=2 \varphi(a)$. We see that $2 \mid a$. Therefore, write $a=2^{k} \cdot m$ with $k \geq 1$ and $m$ odd. Then, by a known property of the $\varphi$-function, we have $\varphi(a)=\varphi\left(2^{k}\right) \varphi(m)=2^{k-1} \cdot \varphi(m)$, so the equation becomes $2^{k} \cdot m=2 \cdot 2^{k-1} \cdot \varphi(m)$, or $m=\varphi(m)$. This implies $m=1$. Therefore, $a=b=2^{k}$, and the equation indeed holds.

We conclude that the solutions are: $(1, p)$ and $(p, 1)$ for all prime numbers $p$, and $\left(2^{k}, 2^{k}\right)$ for all positive integers $k$.