We claim the answer are $n\in\{1,3,5,6\}$. First assume $n$ is a square. Clearly $n=1$ works, let $n\ge 4$. Then $\tau(n)=2k+1$ for some $k$ and $d_1<d_2<\cdots<d_{2k+1}$ be divisors of $n$, where $d_{k+1}=\sqrt{n}\in\mathbb{N}$. We then have $\textstyle \sum_{i\le 2k+1}d_i = (2k+1)\sqrt{n}$. Notice that the divisors of $n$ are paired: for each $1\le i\le k+1$, $d_id_{2k+2-i} = n$. Using AM-GM inequality, \[ \frac{d_i+d_{2k+2-i}}{2} \ge 2\sqrt{d_id_{2k+2-i}} = 2\sqrt{n}\implies \sigma(n) \ge (2k+1)\sqrt{n}. \]Hence, we must retain equality everywhere. But this is possible only when $n=1$ (as $n=d_1^2=1$). Now assume $n$ is not a square. Then $\tau(n)$ is even, set $\tau(n)=2k$ and let $d_1<\cdots<d_{2k}$ be the divisors of $n$. Further, let $t^2<n<(t+1)^2$ thus $\lceil \sqrt{n}\rceil = t+1$. We now consider the sums $S_i = d_i+d_{2k+1-i}$ for $1\le i\le k$. It is not hard to see that $S_1>S_2>\cdots>S_k$ as $d_id_j<n$ for $1\le i\le j\le k$. Now, $d_k+d_{k+1}\ge 2\sqrt{n}>2t\implies d_k+d_{k+1}\ge 2t+1$. From here, we get that $S_i \ge 2t+(k+1-i)$ for $1\le i\le k$. Summing up, we get \[ 2kt+2k=\sigma(n) = \sum_{i\le k}S_i \ge 2tk + \frac{k(k+1)}{2} \implies 3\ge k. \]We now inspect the remaining cases. Assume $k=3$. Then $n+1=k+2t = 2t+3\implies n=2t+2$. Moreover, $t^2<n$, implying $(t-1)^2<3$. Thus $t\in\{1,2\}$. An easy inspection reveals that this brings no solutions. Assume now $k=2$. Then $n=pq$ for some primes $p<q$. This yields \[ (p+1)(q+1) \le 4(\sqrt{pq}+1) \implies 4\sqrt{pq}+4 > pq+2\sqrt{pq}+1 \implies t^2-2t-3<0 \]for $t=\sqrt{pq}$. Easily, $pq<9$. Inspecting we recover $n=6$ (as $\tau(n)=4)$. Finally, let $n=p$ is a prime. Then $(p+1)<2\sqrt{p}+2$. Solving, we get $p\le 5$. This brings $n=3,5$; completing the proof.

Notice that $$\begin{aligned} \sigma(n) -\tau(n) \lceil {\sqrt{n}} \rceil &=\frac 12\sum_{d\mid n}\left(d+\frac nd\right)-\sum_{d\mid n}\left(\sqrt n+(\lceil {\sqrt{n}} \rceil -\sqrt n)\right)\\ &=\frac 12\sum_{d\mid n}\left(d-2\sqrt n+\frac nd\right)-\sum_{d\mid n}(\lceil {\sqrt{n}} \rceil -\sqrt n)\\ &=\frac 12\sum_{d\mid n}\left(\sqrt d-\sqrt{\frac nd}\right)^2-\tau (n)(\lceil {\sqrt{n}} \rceil -\sqrt n)\\ &=0, \end{aligned}$$we have $$\begin{aligned} (\sqrt n-1)^2 &=\frac12\left(\left(\sqrt 1-\sqrt{\frac n1}\right)^2+\left(\sqrt n-\sqrt{\frac nn}\right)^2\right)\\ &\leqslant\frac 12\sum_{d\mid n}\left(\sqrt d-\sqrt{\frac nd}\right)^2\\ &=\tau (n)(\lceil {\sqrt{n}} \rceil -\sqrt n)\\ &<2\sqrt n. \end{aligned}$$Therefore $$n-2\sqrt n+1<2\sqrt n\Rightarrow n-4\sqrt n+1<0\Rightarrow n<\left(\frac{4+\sqrt{16-4}}2\right)^2=7+4\sqrt 3<14.$$Try all positive integers $n\leq 13$ and we know that only $\boxed{n=1,3,5,6}$ satisfies$.\blacksquare$